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IB 


THE 


NATIONAL    ARITHMETIC, 


ON  THE 


INDUCTIVE    SYSTEM, 


COMBINING  THE 


ANALYTIC  AND  SYNTHETIC  METHODS; 


FORMING  A  COMPLETE 


COURSE  OF  HIGHER  ARITHMETIC. 


Bt  benjamin  greenleaf,  am. 

AUTHOR   OF    THE    "COMMON    SCHOOL   ARITHMETIC,"    "  ALGEBRA,"    ETC 


NEW  ELECTROTYPE  EDITION, 

WITH     ADDITIONS     AND     IMPROVEMENTS. 


BOSTON: 
PUBLISHED    BY  ROBERT    S.  DAYIS  &   CO. 

NEW  YORK  :     D.    APPLETON   &    CO.,    AND   MASON   BROTHERS. 

PHILADELPHIA  :     J.   B.   LIPPINCOTT    AND    COMPANY. 

CHICAGO  :    KEEN  AND  LEE. 

1858. 


Entered  according  to  Act  of  Congress,  in  the  year  1847,  by 

BENJAMIN    GREENLEAF, 

In  the  Clerk's  Office  of  the  District  Court  of  the  District  of  Massachusetts. 


Entered  according  to  Act  of  Congress,  in  the  year  1857,  by 

BENJAMIN    GREENLEAF, 

In  the  Clerk's  Office  of  the  District  Court  of  the  District  of  Massachusetts. 


GBEENLEAF'S    SERIES    OF    MATHEMATICS. 

1.  PRIMARY  ARITHMETIC ;  Ok,  MENTAL  ARITHMETIC,  upon  the  Inductive 
Plan  ;  designed  for  Primary  Schools.    Improved  electrotype  edition.    72  pp. 

2.  INTELLECTUAL  ARITHMETIC  |  Ok,  HIGHER  MENTAL  ARITHMETIC,  upon 
the  Inductive  Plan  ;  designed  for  Common  Schools  and  Academies.    Improved  edition. 

3.  COMMON  SCHOOL  ARITHMETIC  ;  Or,  INTRODUCTION  TO  THE  NATIONAL 
ARITHMETIC.    Improved  stereotype  edition.    324  pp. 

4.  HIGHER  ARITHMETIC  ;  Or,  THE  NATIONAL  ARITHMETIC,  being  a  com- 
plete course  of  Higher  Arithmetic,  for  advanced  scholars  in  Common  Schools  and  Acade- 
mies.   New  electrotype  edition,  with  additions  and  improvements.    444  pp. 

5.  PRACTICAL  TREATISE  ON  ALGEBRA,  for  Academies  and  High  Schools,  and 
for  advanced  Students  in  Common  Schools.    Improved  stereotype  edition.    360  pp. 

6.  ELEMENTS  OF  GEOMETRY,  for  Academies  and  High  Schools,  and  for  Advanced 
Students  in  Common  Schools.     [In  preparation,  and  will  soon  be  published.] 

COMPLETE  KEYS  TO  THE  INTRODUCTION,  AND  NATIONAL  ARITHME- 
TIC, AND  THE  PRACTICAL  TREATISE  ON  ALGEBRA,  containing  Solutions  and 
Explanations,  for  Teachers  only.    In  3  volumes. 


EDUCATIOU  LIBH# 


O3  Two  editions  of  the  National  Arithmetic,  and  also  of  the  Common  School  Arith- 
metic, one  containing  the  answers  to  the  examples,  and  the  other  without  them,  are  pnb- 
lished.    Teachers  are  requested  to  state  in  their  orders  which  edition  they  prefer. 


ELECTROTYPED  AND  PRINTED   BY  METCALE  &   CO.,   CAMBRIDGE,   MASS. 


PREFACE. 


The  National  Arithmetic  was  first  presented  to  the  American  pub- 
lic in  1835.  The  generous  favor  with  which  it  was  received  assured 
the  author  that  he  had  not  misunderstood  the  wants  of  the  public  in 
the  department  of  arithmetical  instruction,  and  that  his  labors  had,  to 
a  considerable  extent,  supplied  those  wants. 

During  the  ten  years  following,  increased  attention  was  given  to 
the  subject  of  popular  education,  and  great  improvements  were  made 
in  methods  of  imparting  knowledge.  Accomplished  teachers  soon 
began  to  demand  a  work  on  Arithmetic,  which  should  embody  the 
numerous  improvements  which  had  enriched  that  science.  In  re- 
sponse to  a  demand  so  reasonable,  the  author  was  induced,  in  1847, 
to  prepare  a  revised  and  enlarged  edition  of  the  National  Arithmetic. 
Aided  by  important  suggestions  from  eminent  teachers,  and  directly 
assisted  by  gentlemen  intimately  acquainted  with  arithmetical  sci- 
ence, he  was  enabled  to  produce  a  work  which,  up  to  the  present 
time,  has  been  steadily  increasing  in  public  favor. 

The  last  ten  years  have  formed  a  period  of  unprecedented  activity 
in  all  that  relates  to  the  interests  of  education.  The  numerous  Arith- 
metics which,  within  this  period,  have  become  candidates  for  popular 
patronage,  afford  ample  evidence  that  the  department  of  knowledge 
to  which  they  relate  has  meanwhile  received  its  share  of  attention. 
Vigorous  emulation  among  authors  and  publishers  has  produced  thor- 
ough investigation,  careful  preparation,  and  valuable  results. 

The  author  of  this  work,  wishing,  if  possible,  to  keep  pace  with  the 
rapid  march  of  improvement,  has  again  thoroughly  revised,  rewritten, 
and  considerably  enlarged  it.  The  results  of  a  long  experience  as  a 
mathematical  instructor,  and  the  suggestions  of  many  distinguished 
teachers  of  the  present  day,  are  embodied  in  this  volume. 

In  preparing  this  as  well  as  the  former  editions  of  his  National 
Arithmetic,  the  author  has  regarded  the  end  to  be  sought  in  the  study 
of  Arithmetic  as  twofold,  —  a  practical  knowledge  of  numbers,  and  the 
discipline  of  the  mind.  With  reference  to  the  former,  he  has  endeav- 
ored to  present  methods  which  are  brief,  accurate,  and  especially 
adapted  to  the  wants  of  business  life ;  with  reference  to  the  latter,  he 
has  aimed  to  give  a  clear  and  logical  analysis  of  every  operation,  from 
the  simplest  to  the  most  involved. 

The  author  adheres  to  his  opinion  long  since  advanced,  in  relation 

M577021 


iv  PREFACE. 

to  the  value  of  rules  in  an  arithmetical  treatise.  It  is  not  an  easy- 
thing  for  the  experienced  teacher  to  express  in  the  most  concise 
and  accurate  language  the  method  of  solving  a  problem.  Much  less 
can  such  an  expression  be  given  by  the  untrained  scholar.  Now,  as 
precision  in  thought  is  essentially  aided  by  precision  in  language,  it  is 
deemed  expedient  to  furnish  the  scholar  with  rules  which  shall  state 
in  the  fewest  and  clearest  words  the  results  of  previous  logical  induc- 
tions. Moreover,  when  an  intricate  reasoning  process  may  have  been 
forgotten  and  cannot  readily  be  recalled,  the  brief  form  of  words  im- 
pressed upon  the  memory  in  one's  youth  may  oftentimes  enable  him 
in  after  life  to  perform  an  important  mathematical  operation  in  which 
he  must  otherwise  have  failed. 

It  will  be  observed,  that,  while  the  author  has  expressed  in  rules  his 
modes  of  operating,  he  has  in  every  case  first  given  the  analysis  upon 
which  each  rule  is  based. 

The  author  flatters  himself  that  the  present  edition  of  the  National 
Arithmetic  embraces  many  improvements  on  former  editions.  He  has 
endeavored  to  present  clearer  definitions,  more  rigid  analyses,  and 
briefer  and  more  accurate  rules.  While  almost  every  topic  included 
in  earlier  editions  has  been  treated  in  a  more  elaborate  and  compre- 
hensive manner,  this  volume  comprises  a  large  amount  of  new  matter, 
which  it  is  believed  will  be  found  useful  in  business. 

On  comparing  this  with  preceding  editions,  teachers  will  find  exten- 
sive additions  and  improvements  under  the  heads  of  Numeration, 
Addition,  and  the  other  fundamental  rules,  Properties  of  Numbers, 
Fractions,  Ratio,  Percentage,  Notes  and  Banking,  Roots,  etc.  Among 
the  new  material  will  be  discovered  methods  of  finding  the  greatest 
common  divisor  and  the  least  common  multiple  of  fractions,  of  re- 
ducing fractions  to  a  common  numerator,  of  contracting  the  operations 
in  the  multiplication  and  division  of  decimal  fractions,  of  reducing 
continued  fractions,  of  averaging  accounts,  of  alligating,  of  extracting 
roots  to  any  degree,  and  of  reducing  numbers  from  one  system  of  no- 
tation to  another. 

Especial  attention  is  invited  to  the  section  on  averaging  accounts,  — 
a  subject  rarely  taught  in  schools,  though  of  great  importance  in  the 
counting-room,  —  to  the  manner  of  treating  the  roots,  and  to  the  many 
new  problems  which  will  be  found  in  all  parts  of  the  book. 

In  closing  these  remarks,  the  author  desires  to  tender  his  hearty 
thanks  to  many  teachers  who  have  favored  him  with  valuable  sug- 
gestions; and  to  acknowledge  in  an  especial  manner  his  indebted- 
ness to  Mr.  H.  B.  Maglathlin,  who  has  been  constantly  associated 
with  him  in  making  this  revision,  and  to  whose  accurate  scholarship 
and  sound  judgment  the  value  of  the  work  is  largely  due. 

Bradford,  Mass.,  May  30,  1857. 


CONTENTS. 


Page 

INTRODUCTION 7 

DEFINITIONS 13 

SIGNS 14 

AXIOMS 15 

NOTATION 16 

Roman  Notation 16 

Arabic  Notation 18 

NUMERATION 20 

French  Numeration      ...  20 

English  Numeration      ...  23 

ADDITION  ........  24 

SUBTRACTION 30 

MULTIPLICATION 35 

DIVISION 44 

GENERAL  PRINCIPLES  AND 

APPLICATIONS 53 

Cancellation 54 

Contractions  in  Multiplication  56 

Contractions  in  Division    .     .  62 

PROBLEMS 66 

UNITED  STATES  MONEY.  .  70 

Reduction  of  United  States 

Money 72 

Addition    of    United    States 

Money 73 

Subtraction  of  United  States 

Money 74 

Multiplication  of  United  States 

Money 75 

Division    of    United     States 

Money     .    .  • 76 

General  Principles  and  Ap- 
plications        78 

Analysis  by  Aliquot  Parts     .  78 
1* 


Page 
Bills 83 

Ledger  Accounts      ....    87 
COMPOUND  NUMBERS.    .    .    88 
Reduction  of  CompoundNum- 

bers 88 

Addition  of  Compound  Num- 
bers      Ill 

Subtraction     of     Compound 

Numbers.     .*....  116 
Multiplication  of  Compound 

Numbers 119 

Division  of  Compound  Num- 
bers      123 

PRINCIPLES  AND  APPLICA- 
TIONS   126 

Difference  between  Dates .  .  126 
Difference  of  Latitude  .  .  .  129 
Difference  of  Longitude  .  .  130 
Longitude  and  Time  .  .  .  132 
PROPERTIES  OF  NUMBERS  .  137 

Definitions 137 

Prime  Numbers 139 

Factoring 141 

Divisibility  of  Numbers  .  .  142 
Greatest  Common  Divisor  .  146 
Least  Common  Multiple  .  .149 
COMMON  FRACTIONS  .  .  .  154 
Reduction  of  Common  Frac- 
tions     155 

A  Common  Denominator  .     .  159 
Addition  of  Common  Frac- 
tions     161 

Subtraction  of  Common  Frac- 
tions     164 

Multiplication    of    Common 
Fractions 169 


VI 


CONTENTS. 


Division  of  Common    Frac- 
tions     173 

Complex  Fractions  .    .     .    .177 
Denominate  Fractions  .     .    .185 
DECIMAL  FRACTIONS  ...  197 
Notation  and  Numeration  of 

Decimals 198 

Addition  of  Decimals  .  .  .  200 
Subtraction  of  Decimals  .  .  202 
Multiplication  of  Decimals  .  203 
Division  of  Decimals  .  .  .  206 
Reduction  of  Decimals  .  .  210 
CIRCULATING  DECIMALS  .  217 
Reduction  of  Repetends  .  .  219 
Transformation  of  Repetends  222 
Addition  of  Circulating  Deci- 
mals     224 

Subtraction    of    Circulating 

Decimals 225 

Multiplication  of  Circulating 

Decimals 225 

Division  of  Circulating  Deci- 
mals     226 

CONTINUED  FRACTIONS  .    .  227 

RATIO 229 

Reduction  and  Comparison  of 

Ratios 232 

Analysis  by  Ratio    ....  234 

PROPORTION 237 

Simple  Proportion  .  .  .  .238 
Compound  Proportion  .  .  .  244 
Conjoined  Proportion   .     .    .  248 

PERCENTAGE 251 

Interest 259 

Promissory  Notes  .  .  .  .272 
Partial  Payments  ....  273 
Compound  Interest  .  .  .  .279 
Discount  and  Present  Worth   286 

Banking 289 

Stocks 295 

Brokerage  and  Commission  .  299 
Account  of  Sales .     .    .     .    .  302 

Profit  and  Loss 303 

PARTNERSHIP,   OR    COMPA- 
NY BUSINESS 308 

Bankruptcy 313 


Taxes 314 

General  Average 818 

EQUATION  OF  PAYMENTS  .  320 
Averaging  of  Accounts  .  .  324 
Accounts  of  Storage     .    .     .  332 

INSURANCE 335 

Fire  and  Marine  Insurance    .  335 

Life  Insurance 337 

CUSTOM-HOUSE  BUSINESS  .  341 

COINS  AND  CURRENCIES     .  345 

Reduction  of  Currencies   .     .  347 

EXCHANGE 349 

Inland  Bills 353 

Foreign  Bills 354 

Arbitration  of  Exchanges .    .358 

ALLIGATION 363 

Alligation  Medial  ....  363 
Alligation  Alternate      .     .     .  364 

INVOLUTION 369 

EVOLUTION 371 

Extraction  of  the  Square  Root  373 
Extraction  of  the  Cube  Root    378 
Extraction  of  any  Root     .     .384 
APPLICATIONS  OF  POWERS 

AND   ROOTS 388 

PROGRESSION,  OR  SERIES  .  395 
Arithmetical  Progression  .  .  395 
Geometrical  Progression   .     .  400 

ANNUITIES 405 

PERMUTATIONS   AND    COM- 
BINATIONS   411 

ANALYSIS  BY  POSITION.    .  413 
SCALES  OF  NOTATION    .    .  417 

DUODECIMALS 420 

Addition  and  Subtraction  of 

Duodecimals 421 

Multiplication  of  Duodecimals  421 
Division  of  Duodecimals   .     .  423 

MENSURATION 425 

Definitions 425 

Mensuration  of  Surfaces  .  .  425 
Mensuration  of  Solids  .  .  .433 
Mensuration  of  Lumber  .  .437 
Gauging  of  Casks  ....  439 
Tonnage  of  Vessels  ....  439 


INTRODUCTION. 


HISTORY    OF     ARITHMETIC. 

It  is  difficult  to  determine  who  was  the  inventor  of  Arithmetic,  or 
in  what  age  or  among  what  people  it  originated.  In  ordinary  history, 
we  find  the  origin  of  the  science  attributed  by  some  to  the  Greeks,  by 
some  to  the  Chaldeans,  by  some  to  the  Phoenicians,  by  Josephus  to 
Abraham,  and  by  many  to  the  Egyptians.  The  opinion,  however, 
which  modern  investigations  have  rendered  most  probable,  is,  that 
Arithmetic,  properly  so  called,  is  of  Indian  origin,  —  that  is,  that  the 
science  received  its  first  definite  form,  and  became  the  germ  of  mod- 
ern Arithmetic,  in  the  regions  of  the  East. 

It  is  evident,  from  the  nature  of  the  case,  that  some  knowledge  of 
numbers  and  of  the  art  of  calculation  was  necessary  to  men  in  the 
earliest  periods  of  society,  since  without  this  they  could  not  have  per- 
formed the  simplest  business  transactions,  even  such  as  are  incidental 
to  an  almost  savage  state.  The  question,  therefore,  as  to  the  invention 
of  Arithmetic,  deserves  to  be  considered  only  as  it  respects  the  origin 
of  the  science  as  we  now  have  it,  and  which,  as  all  scholars  admit,  has 
reached  a  surprising  degree  of  perfection.  In  this  sense  the  honor  of 
the  invention  must  be  awarded  to  the  Hindoos. 

The  history  of  the  various  methods  of  Notation,  or  the  different 
means  by  which  numbers  have  been  expressed  by  signs  or  characters, 
is  one  of  much  interest  to  the  advanced  and  curious  scholar ;  but  the 
brevity  of  this  sketch  allows  us  barely  to  touch  upon  it  here.  Among 
the  ancient  -  nations  which  possessed  the  art  of  writing,  it  was  a  natu- 
ral and  common  device  to  employ  letters  to  denote  what  we  express 
by  our  numeral  figures.  Accordingly  we  find,  that,  with  the  Hebrews 
and  Greeks,  the  first  letter  of  their  respective  alphabets  was  used  for 
1,  the  second  for  2,  and  so  on  to  the  number  10,  —  the  latter,  how- 
ever, inserting  one  new  character  to  denote  the  number  6,  and  evi- 
dently in  order  that  their  notation  might  coincide  with  that  of  the 
Hebrews,  the  sixth  letter  of  the  Hebrew  alphabet  having  no  corre- 
sponding one  in  the  Greek. 


8  INTRODUCTION. 

The  Komans,  as  is  well  known,  employed  the  letters  of  their  alpha- 
bet as  numerals.  Thus  I  denotes  1 ;  Y,  5  ;  X,  10  ;  L,  50  ;  C,  100  ; 
D,  500  ;  and  M,  1,000.  The  intermediate  numbers  were  expressed 
by  a  repetition  of  these  letters  in  various  combinations  ;  as,  II  for  2  ; 
VI  for  6 ;  XV  for  15 ;  IV  for  4  ;  IX  for  9,  &c.  They  frequently  ex- 
pressed any  number  of  thousands  by  the  letter  or  letters  denoting  so 
many  units,  with  a  line  drawn  above ;  thus,  V,  5,000 ;  VI,  6,000 ;  X, 
10,000  ;  L,  50,000  ;  C,  100,000;  M,  1,000,000. 

In  the  classification  of  numbers,  as  well  as  in  the  manner  of  ex- 
pressing them,  there  has  been  a  great  diversity  of  practice.  While 
we  adopt  the  decimal  scale  and  reckon  by  tens,  the  aborigines  of 
Mexico,  according  to  Humboldt,  and  some  of  the  early  nations  of 
Europe,  adopted  the  vicenary,  reckoning  by  twenties ;  some  of  the 
Indian  tribes,  and  several  of  the  African  tribes,  use  the  quinary, 
reckoning  by  fives ;  and  the  Chinese  for  more  than  4,000  years 
have  used  the  Unary,  reckoning  by  twos.  The  adoption  of  one  or 
another  of  these  scales  has  been  so  general,  that  they  have  been  re- 
garded as  natural,  and  accounted  for  by  referring  them  to  a  common 
and  natural  cause.  The  reason  for  assuming  the  binary  scale  prob- 
ably lay  in  the  use  of  the  two  hands,  which  were  employed  as  counters 
in  computing ;  that  for  employing  the  quinary,  in  a  similar  use  of  the 
five  fingers  on  either  hand ;  while  the  decimal  and  vicenary  scales 
had  respect,  the  former  to  the  ten  fingers  on  the  two  hands,  and  the 
latter  to  the  ten  fingers  combined  with  the  ten  toes  on  the  naked  feet, 
which  were  as  familiar  to  the  sight  of  a  rude,  uncivilized  people  as 
their  fingers.  -»-  It  is  an  interesting  circumstance,  that  in  the  common 
name  of  our  numeral  figures,  digits  (digiti)  or  fingers,  we  preserve  a 
memento  of  the  reason  why  ten  characters  and  our  present  decimal 
scale  of  numeration  were  originally  adopted  to  express  all  numbers, 
even  of  the  highest  order. 

It  is  now  almost  universally  admitted,  that  our  present  numeral 
characters,  and  the  method  of  estimating  their  value  in  a  tenfold  ratio 
from  right  to  left,  have  decided  advantages  over  all  other  systems, 
both  of  notation  and  numeration,  that  have  ever  been  adopted. 
There  have  been  those,  as  Leibnitz  and  De  Lagni,  who  have  advo- 
cated the  binary  scale ;  a  few,  with  Claudius  Ptolemy,  have  claimed 
advantages  for  the  sexagenary  scale,  or  that  by  sixty ;  and  there  are 
those  who  think  that  a  duodecimal  scale,  and  the  use  of  twelve  numeral 
figures  instead  of  ten,  would  afford  increased  facility  for  rapid  and  ex- 
tensive calculations ;  but  most  mathematicians  are  satisfied  with  the 
present  number  of  numerals  and  the  scale  of  numeration,  which  have 
attained  an  adoption  all  but  universal. 

It  was  long  supposed,  that  for  our  modern  Arithmetic  the  world  is 
indebted  to  the  Arabians.     But  this,  as  we  have  seen,  is  not  the 


HISTORY   OF   ARITHMETIC.  9 

case.  The  Hindoos  at  least  communicated  a  knowledge  of  it  to  the 
Arabians,  and,  as  we  are  not  able  to  trace  it  beyond  the  former  people, 
they  must  have  the  honor  of  its  invention.  They  do  not,  however, 
claim  this  honor,  but  refer  it  to  the  Divinity,  declaring  that  the  inven- 
tion of  nine  figures,  with  device  of  place,  is  to  be  ascribed  to  the  benefi- 
cent Creator  of  the  universe. 

But  though  the  invention  of  modern  Arithmetic  is  to  be  ascribed  to 
the  Hindoos,  the  honor  of  introducing  it  into  Europe  belongs  unques- 
tionably to  the  Arabians.  It  was  they  who  took  the  torch  from  the 
East  and  passed  it  along  to  the  West.  The  precise  period,  however, 
at  which  this  was  done,  it  is  not  easy  to  determine.  It  is  evident,  that 
our  numeral  characters  and  our  method  of  computing  by  them  were 
in  use  among  the  Arabians  about  the  beginning  of  the  eighth  century, 
when  they  invaded  Spain,  and  it  is  probable  that  a  knowledge  of  them 
was  soon  afterwards  communicated  to  the  inhabitants  of  Spain,  and 
gradually  to  those  of  the  other  European  countries. 

It  is  said,  that  the  celebrated  Gerbert,  afterward  Pope  Sylvester  II., 
returning  to  France  from  Spain,  where  he  had  been  to  acquire  a 
knowledge  of  the  Arabic  or  Indian  notation,  about  the  year  970,  in- 
troduced it  among  the  French. 

About  the  middle  of  the  eleventh  century  it  is  supposed  to  have 
been  introduced  into  England  by  John  of  Basingstoke,  Archdeacon  of 
Leicester. 

The  Arabic  characters,  having  been  first  used  by  astronomers,  be- 
came circulated  over  Europe  in  their  almanacs ;  but  do  not  seem  to 
have  secured  general  adoption  in  Europe  earlier  than  the  twelfth  or 
thirteenth  century. 

The  science  of  Arithmetic,  like  all  other  sciences,  was  very  limited 
and  imperfect  at  the  beginning,  and  the  successive  steps  by  which  it 
has  reached  its  present  extension  and  perfection  have  been  taken  at 
long  intervals  and  among  different  nations.  It  has  been  developed 
by  the  necessities  of  business,  by  the  strong  love  of  certain  minds  for 
mathematical  science  and  numerical  calculation,  and  by  the  call  for  its 
higher  offices  by  other  sciences,  especially  that  of  Astronomy.  In  its 
progress,  we  find  that  the  Arabians  discovered  the  method  of  proof  by 
casting  out  the  9's,  and  that  the  Italians  early  adopted  the  practice  of 
separating  numbers  into  periods  of  six  figures,  for  the  purpose  of  enu- 
meration. To  facilitate  the  process  of  multiplication,  this  latter  people 
also  introduced,  probably  from  the  writings  of  Boethius,  the  Multipli- 
cation Table  of  Pythagoras. 

The  invention  of  the  Decimal  Fraction  was  a  great  step  in  the  ad- 
vancement of  arithmetical  science,  and  the  honor  of  it  has  generally 
been  given  to  John  Muller,  commonly  called  Regiomontanus,  about 
the  year  1464.     It  appears,  however,  that  Stevinus,  in  1582,  wrote 


10  INTRODUCTION. 

the  first  express  treatise  on  the  subject.  The  credit  of  first  using  the 
decimal  point,  by  which  the  invention  became  permanently  available, 
is  given  by  Dr.  Peacock  to  Napier,  the  inventor  of  Logarithms ;  but 
De  Morgan  says,  that  it  was  used  by  Richard  Witt  as  early  as  1613, 
while  it  is  not  shown  that  Napier  used  it  before  1617.  Circulating 
Decimals  received  but  little  attention  till  the  time  of  Dr.  Wallis,  the 
author  of  the  Arithmetic  of  Infinites.  Dr.  Wallis  died  at  Oxford,  in 
1703. 

The  greatest  improvement  which  the  art  of  computation  ever  re- 
ceived was  the  invention  of  Logarithms,  the  honor  of  which  is  un- 
questionably due  to  Baron  Napier,  of  Scotland,  about  the  end  of  the 
sixteenth  or  the  commencement  of  the  seventeenth  century. 

The  oldest  treatises  on  Arithmetic  now  known  are  the  7th,  8th, 
9th,  and  10th  books  of  Euclid's  Elements,  in  which  he  treats  of  pro- 
portion and  of  prime  and  composite  numbers.  These  books  are  not 
contained  in  the  common  editions  of  the  great  geometer,  but  are 
found  in  the  edition  by  Dr.  Barrow,  the  predecessor  of  Sir  Isaac 
Newton  in  the  mathematical  chair  at  Cambridge.  Euclid  flourished 
about  300  B.  C. 

A  century  later,  Eratosthenes  invented  a  method,  which  is  known 
as  his  "  sieve,"  for  separating  prime  numbers  from  others. 

The  next  writer  on  Arithmetic  mentioned  in  history  is  Nico- 
machus,  the  Pythagorean,  who  wrote  a  treatise  relating  chiefly  to 
the  distinctions  and  divisions  of  numbers  into  classes,  as  plain,  solid, 
triangular,  &c.     He  is  supposed  to  have  lived  near  the  Christian  era. 

About  the  middle  of  the  fourth  century  lived  Diophantus,  a  cele- 
brated mathematician,  who,  besides  being  the  first  known  author  on 
the  subject  of  Algebra,  composed  thirteen  books  on  Arithmetic,  six  of 
which  are  still  extant. 

The  next  writer  of  note  is  Boethius,  the  Roman,  who,  however, 
copied  most  of  his  work  from  Nicomachus.  He  lived  at  the  begin- 
ning of  the  sixth  century,  and  is  the  author  of  tljp  well-known  work 
on  the  Consolation  of  Philosophy. 

The  next  writer  of  eminence  on  the  subject  is  Jordanus,  of  Namur, 
who  wrote  a  treatise  about  the  year  1200,  which  was  published  by 
Joannes  Faber  Stapulensis  in  the  fifteenth  century,  soon  after  the 
invention  of  printing. 

The  author  of  the  first  printed  treatise  on  Arithmetic  was  Pacioli, 
or,  as  he  is  more  frequently  called,  Lucas  de  Burgo,  an  Italian  monk, 
who  in  1484  published  his  great  work  entitled  Summa  de  Arithmetica^ 
&c,  in  which  our  present  numerals  appear  under  very  nearly  their 
modern  form. 

In  1522,  Bishop  Tonstall  published  a  work  on  the  Art  of  Computa- 
tion, in  the  Dedication  of  which  he  says,  that  he  was  induced  to  study 


HISTORY   OF   ARITHMETIC.  11 

Arithmetic  to  protect  himself  from  the  frauds  of  money-changers  and 
stewards,  who  took  advantage  of  the  ignorance  of  their  employers.  In 
his  preparation  for  this  work,  he  professes  to  have  read  all  the  books 
which  had  been  published  on  this  subject,  adding,  also,  that  there  was 
hardly  any  nation  which  did  not  possess  such  books. 

About  the  year  1540,  Robert  Eecord,  Doctor  in  Physic,  printed 
the  first  edition  of  his  famous  Arithmetic,  which  was  afterward  aug- 
mented by  John  Dee,  and  subsequently  by  John  Mellis,  and  which 
did  much  to  advance  the  science  and  practice  of  Arithmetic  in  Eng- 
land in  its  early  stages.  This  work,  which  is  now  quite  a  curiosity, 
effectually  destroys  the  claim  to  originality  in  some  things  of  which 
authors  much  more  modern  have  obtained  the  credit.  In  it  we  find 
the  celebrated  case  of  a  will,  which  we  have  in  the  Miscellaneous 
Questions  of  Webber  and  Kinne,  and  which,  altered  in  language  and 
the  time  of  making  the  testament,  is  the  2nd  Miscellaneous  Question 
in  the  present  work.  This  question  is,  by  his  own  confession,  older 
than  Record,  and  is  said  to  have  been  famous  since  the  days  of  Lucas 
de  Burgo.  In  Record  it  occurs  under  the  "  Rule  of  Fellowship." 
Record  was  the  author  of  the  first  treatise  on  Algebra  in  the  English 
language. 

In  1556,  a  complete  work  on  Practical  Arithmetic  was  published  by 
Nicolas  Tartaglia,  an  Italian,  and  one  of  the  most  eminent  mathema- 
ticians of  his  time. 

From  the  time  of  Record  and  Tartaglia,  works  on  Arithmetic  have 
been  too  numerous  to  mention  in  an  ordinary  history  of  the  science. 
De  Morgan,  in  his  recent  work  (Arithmetical  Books),  has  given  the 
names  of  a  large  number,  with  brief  observations  upon  them,  and  to 
this  the  inquisitive  student  is  referred  for  further  information  in  re- 
gard both  to  writers  and  books  on  this  subject  since  the  invention  of 
Printing.  It  is  remarkable  that  De  Morgan  knew  next  to  nothing 
of  any  American  works  on  Arithmetic.  He  mentions  the  "  Ameri- 
can Accountant"  by  William  Milns,  New  York,  1797,  and  gives  the 
name  of  Pike  (probably  Nicholas  Pike)  among  the  names  of  which 
he  had  heard  in  connection  with  the  subject.  Of  the  compilation  of 
Webber  and  the  original  work  of  Walsh,  he  seems  to  have  been  en- 
tirely ignorant. 

The  various  signs  or  symbols,  which  are  now  so  generally  used  to 
abridge  arithmetical  as  well  as  algebraical  operations,  were  introduced 
gradually,  as  necessity  or  convenience  taught  their  importance.  The 
earliest  writer  on  Algebra  after  the  invention  of  printing  was  Lucas 
de  Burgo,  above  mentioned,  and  he  uses  p  for  plus  and  m  for  minus, 
and  indicates  the  powers  by  the  first  two  letters,  in  which  he  is  fol- 
lowed by  several  of  his  successors.  After  this,  Steifel,  a  German, 
who  in  1544  published  a  work  entitled  Arithmetica  Integra,  added 


12  INTRODUCTION. 

considerably  to  the  use  of  signs,  and,  according  to  Dr.  Hutton,  was 
the  first  who  employed  -f-  and  —  to  denote  addition  and  subtraction. 
To  denote  the  root  of  a  quantity  he  also  used  our  present  sign  tJlx 
originally  r,  the  initial  of  the  word  radix,  root.  The  sign  ==,  to  denote 
equality,  was  introduced  by  Record,  the  above-named  English  mathe- 
matician, and  for  this  reason,  as  he  says,  that  "  noe  2  thynges  can  be 
moar  equalle,"  namely,  than  two  parallel  lines.  It  is  a  curious  cir- 
cumstance that  this  same  symbol  was  first  used  to  denote  subtraction. 
It  was  also  employed  in  this  sense  by  Albert  Girarde,  who  lived  a 
little  later  than  Record.  Girarde  dispensed  with  the  vinculum  em- 
ployed by  Steifel,  as  in  3+4,  and  substituted  the  parenthesis  (3  +  4), 
now  so  generally  adopted.  The  first  use  of  the  St.  Andrew's  cross, 
X,  to  signify  multiplication,  is  attributed  to  William  Oughtred,  an 
Englishman,  who  in  1631  published  a  work  entitled  Clavis  Mathe- 
matical, or  Key  of  Mathematics. 

It  was  intended  to  notice  several  other  works,  ancient  and  modern, 
but  the  length  to  which  this  sketch  has  already  extended  forbids  it. 

We  had  thought  of  alluding  to  the  ancient  philosophic  Arithmetic, 
and  the  elevated  ideas  which  many  of  the  early  philosophers  had  of 
the  science  and  properties  of  numbers.  But  a  word  must  here  suffice. 
Arithmetic,  according  to  the  followers  of  Plato,  was  not  to  be  studied 
"  with  gross  and  vulgar  views,  but  in  such  a  manner  as  might  enable 
men  to  attain  to  the  contemplation  of  numbers  ;  not  for  the  purpose 
of  dealing  with  merchants  and  tavern-keepers,  but  for  the  improve- 
ment of  the  mind,  considering  it  as  the  path  which  leads  to  the  knowl- 
edge of  truth  and  reality."  These  transcendentalists  considered  per- 
fect numbers,  compared  with  those  which  are  deficient  or  superabun- 
dant, as  the  images  of  the  virtues,  which,  they  allege,  are  equally 
remote  from  excess  and  defect,  constituting  a  mean  between  them ; 
as  in  the  case  of  true  courage,  which,  they  say,  lies  midway  between 
audacity  and  cowardice,  and  of  liberality,  which  is  a  mean  between 
profusion  and  avarice.  In  other  respects,  also,  they  regard  this  anal- 
ogy as  remarkable ;  perfect  numbers,  like  the  virtues,  are  "  few  in 
number  and  generated  in  a  constant  order  ;  while  superabundant  and 
deficient  numbers  are,  like  vices,  infinite  in  number,  disposable  in  no 
regular  series,  and  generated  according  to  no  certain  and  invariable 
law." 

We  conclude  this  brief  sketch  in  the  earnest  hope  that  the  noble 
science  of  numbers  may  erelong  find  some  devoted  friend,  who  shall 
collect,  arrange,  and  bring  within  the  reach  of  ordinary  students, 
much  more  fully  than  we  have  done,  the  scattered  details  of  its  long- 
neglected  history. 


ARITHMETIC. 


DEFINITIONS. 

Article  1,  Quantity  is  anything  that  can  be  increased, 
diminished,  or  measured ;  as  time,  weight,  lines,  surfaces,  and 
solids. 

2.  A  unit  is  a  single  thing  or  quantity  regarded  as  a 
whole. 

3 1  An  abstract  unit  is  one  that  has  no  reference  to  any 
particular  thing  or  quantity. 

4.  A  concrete  unit  is  one  that  has  reference  to  some  par- 
ticular thing  or  quantity. 

5.  A  number  is  an  expression  of  quantity,  representing 
either  a  unit  or  a  collection  of  units. 

6.  An  abstract  number  is  a  number  whose  unit  is  abstract ; 
as,  one,  six,  nine. 

7.  A  concrete  or  denominate  number  is  a  number  whose 
unit  is  concrete ;  as,  one  dollar,  six  pounds,  nine  men. 

8.  A  simple  number  is  a  unit,  or  a  collection  of  units, 
either  abstract  or  concrete,  of  a  single  kind  or  denomination ; 
as,  1,  15,  1  book,  13  dollars. 

9t  The  unit  of  measure  of  any  quantity  is  one  of  the 
same  kind  with  that  by  which  the  quantity  is  measured  or  com- 
pared ;  as,  in  the  abstract  number,  six,  the  abstract  unit  is  that 
of  measure  or  comparison ;  and  in  six  pounds,  the  concrete 
unit,  one  pound,  is  that  of  measure  or  comparison. 

10,  Arithmetic  is  the  science  of  numbers  and  the  art  of 
computing  by  them.  It  treats  of  the  properties  and  relations 
N    2 


14  SIGNS. 

of  numbers,  and  teaches  the  methods  of  applying  the  prin- 
ciples of  the  science  to  practical  purposes. 

11.  An  axiom  is  a  self-evident  truth. 

12.  A  problem  is  a  question  proposed  for  solution,  or 
something  to  be  done. 

13.  An  operation  is  the  process  of  finding,  from  given 
quantities,  others  that  are  required. 

14.  A  sign  is  a  symbol  employed  to  indicate  the  relations 
of  quantities,  or  operations  to  be  performed  upon  them. 

15.  A  rule  is  a  direction  for  performing  an  operation. 

16.  An  example  is  a  particular  application  of  a  general 
principle  or  rule. 

17.  The  principal  or  fundamental  processes  of  arithmetic 
are  Notation  and  Numeration,  Addition,  Subtraction,  Multipli- 
cation, and  Division. 

.  SIGNS. 

18.  The  sign  of  equality,  two  short  horizontal  lines,  =, 
is  read  equal,  or  equal  to,  and  denotes  that  the  quantities  be- 
tween which  it  is  placed  are  equal  to  each  other.  Thus,  12 
inches  =  1  foot,  signifies  that  12  inches  are  equal  to  1  foot. 

19.  The  sign  of  addition,  an  erect  cross,  -\-,  is  read  plus, 
and,  or  added  to,  and  denotes  that  the  quantities  between 
which  it  is  placed  are  to  be  added  together.  Thus,  8  — |—  6 
signifies  that  6  is  to  be  added  to  8. 

20.  The  sign  of  subtraction,  a  short  horizontal  line,  — , 
is  read  minus,  or  less,  and  denotes  that  the  quantity  on  the 
right  of  it  is  to  be  subtracted  from  the  quantity  on  the  left. 
Thus,  8  —  6  signifies  that  6  is  to  be  subtracted  from  8. 

21.  The  sign  of  multiplication,  an  inclined  cross,  X?  is 
read  times,  or  midtiplied  by,  and  denotes  that  the  quantities 
between  which  it  is  placed  are  to  be  multiplied  together. 
Thus,  7x6  signifies  that  7  is  to  be  multiplied  by  6. 

22*  The  sign  of  division,  a  horizontal  line  between  two 
dots,  --,  is  read  divided  by,  and  denotes  that  the  quantity  on 
the  left  of  it  is  to  be  divided  by  that  on  the  right.  Thus, 
42-7-6  signifies  that  42  is  to  be  divided  by  6. 

23.     The  sign  of  aggregation,  a  parenthesis,  (   ),   includ- 


AXIOMS.  15 

ing  several  numbers,  or  a  vinculum, ,  drawn  over  them, 

indicates  that  the  value  of  the  expression  is  to  be  used  as 
a  single  number.  Thus,  (17  -)-  3)  X  5,  indicates  that  the 
sum  of  17  and  3,  or  20,  is  to  be  multiplied  by  5;  and 
12  +(9  —  3)-r-2,  indicates  that  the  difference  between  9 
and  3  divided  by  2,  or  3,  is  to  be  added  to  12. 

AXIOMS. 

24.  Arithmetic,  in  common  with  other  branches  of  the 
mathematics,  is  based  upon  axioms,  few  in  number,  and  uni- 
versally admitted  to  be  so  clearly  true,  that  no  process  of 
reasoning  can  make  them  plainer ;  as, 

1.  If  the  same  quantity,  or  equal  quantities,  be  added  to 
equal  quantities,  the  sums  will  be  equal. 

2.  If  the  same  quantity,  or  equal  quantities,  be  subtracted 
from  equal  quantities,  the  remainders  will  be  equal. 

3.  If  the  same  quantity,  or  equal  quantities,  be  added  to 
unequal  quantities,  the  sums  will  be  unequal. 

4.  If  the  same  quantity,  or  equal  quantities,  be  subtracted 
from  unequal  quantities,  the  remainders  will  be  unequal. 

5.  If  equal  quantities  be  multiplied  by  the  same  quantity, 
or  equal  quantities,  the  products  will  be  equal. 

6.  If  equal  quantities  be  divided  by  the  same  quantity,  or 
equal  quantities,  the  quotients  will  be  equal. 

7.  If  the  same  quantity  be  both  added  to  and  subtracted 
from  another,  the  value  of  the  latter  will  not  be  changed. 

8.  If  a  quantity  be  both  multiplied  and  divided  by  the  same 
quantity,  its  value  will  not  be  changed. 

9.  If  two  quantities  be  equally  increased  or  diminished, 
their  difference  will  not  be  changed. 

10.  Quantities  which  are  equal  to  the  same  quantity  are 
equal  to  each  other. 

11.  Quantities  which  are  like  parts  of  equal  quantities  are 
equal  to  each  other. 

12.  The  whole  of  a  quantity  is  greater  than  any  of  its 
parts. 

13.  The  whole  of  a  quantity  is  equal  to  the  sum  of  all  its 
parts. 


16  NOTATION. 

NOTATION  AND  NUMERATION. 
NOTATION. 

25.  Notation  is  the  process  of  representing  numbers  by 
letters,  figures,  or  other  symbols. 

The  common  methods  of  expressing  numbers  are  three  :  by 
words,  written  or  spoken ;  by  letters,  called  the  Roman  meth- 
od ;  and  by  figures,  called  the  Arabic  method. 

26.  In  common  language,  we  express  numbers  by  the 
terms  one,  two,  three,  four,  five,  six,  seven,  eight,  nine,  ten, 
eleven,  twelve,  thirteen,  fourteen,  fifteen,  sixteen,  seventeen,  eigh- 
teen, nineteen,  twenty,  twenty-one,  etc.,  giving  a  distinct  name  to 
each  unit  as  far  as  ten,  when  we  begin  a  second  ten,  and  pass 
on  to  twenty ;  a  third  ten,  and  pass  on  to  thirty  ;  and  so  on  to 
forty,  fifty,  sixty,  seventy,  eighty,  and  ninety.  Proceeding  thus 
we  reach  ten  tens,  which  we  call  one  hundred,  when  we  begin  a 
second  hundred,  and  pass  to  two  hundred ;  a  third  hundred,  and 
pass  to  three  hundred;  and  so  on  as  far  as  ten  hundred,  which 
we  call  one  thousand.  A  thousand  thousand  we  call  one  mil- 
lion ;  a  thousand  million,  one  billion  ;  a  thousand  billion,  one 
trillion  ;  and  so  on  with  numbers  still  higher. 

Note  1.  —  The  term  eleven  is  a  contraction  of  one  left  after  ten ;  and  twelve, 
of  two  left  after  ten.  Thirteen,  fourteen,  fifteen,  sixteen,  seventeen,  eighteen, 
nineteen,  are  derived  from  three  and  ten,  four  and  ten,  five  and  ten,  etc.  Twen- 
ty, thirty,  forty,  fifty,  sixty,  seventy,  eighty,  and  ninety  are  contractions  of  two 
tens,  three  tens,  four  tens,  etc. 

Note  2.  —  Billion  is  a  contraction  of  the  Latin  his,  twice,  and  million ;  and 
trillion,  of  the  Latin  tres,  three,  and  million.  In  like  manner  from  the  Latin 
numerals,  quatuor,  four ;  quinque,  five ;  sex,  six;  septem,  seven;  octo,  eight; 
novem,  nine ;  decern,  ten;  undecim,  eleven ;  duodecim,  twelve ;  tredecim,  thirteen ; 
quatuordecim,  fourteen ;  quindecim,  fifteen ;  sexdecim,  sixteen ;  septendecim, 
seventeen ;  octodecim,  eighteen  ;  novemdecim,  nineteen  ;  viginti,  twenty,  —  are 
formed  quadrillions,  quintillions,  sextillions,  septillions,  octillions,  nonillions,  decilr 
lions,  undecillions,  duodecillions,  etc. 

Roman  Notation. 

27.  The  Roman  Notation,  so  called  from  its  having 
originated  with  the  ancient  Romans,  employs  in  expressing 
numbers  seven  capital  letters,  viz. : 


NOTATION. 


17 


one, 


V, 

five, 


X, 

ten, 


fifty, 


C, 
one  hundred, 


five  hundred, 


M. 

one  thousand. 


All  intervening  and  succeeding  numbers  are  expressed  by 
use  of  these  letters,  either  in  repetitions  or  combinations.  By 
a  letter  being  written  after  another  denoting  equal  or  less  value, 
the  sum  of  their  values  is  represented ;  as,  II  represents  two  ; 
VI,  six.  By  writing  a  letter  denoting  a  less  value  before  a  let- 
ter denoting  a  greater,  their  difference  of  value  is  represented ; 
as,  IV  represents  four  ;  XL,  forty. 

A  dash  ( — )  placed  over  a  letter  increases  the  value  denoted 
by  the  letter  a  thousand  times;  as,  V  represents  five  thousand; 
IV,  four  thousand. 

Table. 


I      denotes 

I  one. 

XXX  de 

note 

s  thirty. 

II 

a 

two. 

XL 

H 

forty. 

ni 

a 

three. 

L 

a 

fifty. 

IV 

u 

four. 

LX 

Ci 

sixty. 

V 

a 

five. 

LXX 

a 

seventy. 

VI 

a 

six. 

LXXX 

a 

eighty. 

vri 

a 

seven. 

XC 

a 

ninety. 

VIII 

u 

eight . 

C 

a 

one  hundred. 

IX 

a 

nine. 

CI 

a 

one  hundred  and  one 

X 

a 

ten. 

CC 

it 

two  hundred. 

XI 

a 

eleven. 

CCC 

a 

three  hundred. 

XII 

a 

twelve. 

CCCC 

a 

four  hundred. 

XIII 

a 

thirteen. 

D 

a 

live  hundred. 

XIV 

u 

fourteen. 

DC 

a 

six  hundred. 

XV 

u 

fifteen. 

DCC 

a 

seven  hundred. 

XVI 

a 

sixteen. 

DCCC 

a 

eight  hundred. 

XVII 

a 

seventeen. 

DCCCC 

a 

nine  hundred. 

XVIII 

a 

eighteen. 

M 

i( 

one  thousand. 

XIX 

a 

nineteen. 

MD 

a 

fifteen  hundred. 

XX 

a 

twenty. 

MM 

a 

two  thousand. 

XXI 

u 

twenty-one. 

XIX 

a 

nineteen  thousand. 

xxn 

a 

twenty-two. 

M 

a 

one  million. 

XXIII 

U 

twenty-three. 

MM 

a 

two  million. 

Note  1.  —  The  Roman  method  of  Notation  is  now  but  little  used,  except  in 
numbering  sections,  chapters,  and  other  divisions  of  books ;  and  for  indicating 
the  hours  on  the  face  of  clocks,  watches,  or  dials. 

2* 


18  NOTATION. 

Note  2.  —  Formerly  CIO  was  used  to  represent  one  thousand,  and  the  pre- 
fixing of  a  C  and  the  annexing  of  a  0  increased  the  number  denoted  ten  times ; 
thus,  CCIOO  represented  ten  thousand,  and  CCCIOOO,  one  hundred  thou- 
sand. 

Exercises. 

Represent  the  following  numbers  by  letters  :  — 

1.  Forty-nine.  Ans.  XLIX. 

2.  Ninety-seven. 

3.  One  hundred  and  eighty-eight. 

4.  Two  hundred  and  nineteen. 

5.  Six  hundred  and  sixty-three. 

6.  One  thousand  five  hundred  and  six. 

7.  One  thousand  eight  hundred  and  fifty-seven. 

8.  Four  thousand  four  hundred  and  forty-four. 

9.  Eleven  thousand  nine  hundred  and  eleven. 

10.  One  hundred  fifty  thousand  and  fifty. 

11.  One  million  twenty  thousand  and  twenty. 

12.  Three  million  one  hundred  thousand. 

Arabic  Notation. 

28.  Arabic  Notation,  so  called  from  its  having  been 
made  known  through  the  Arabs,  employs  in  expressing  num- 
bers ten  characters  or  figures,  viz. : 

1,       2,        3,        4,      5,       6,        7,  8,        9,  0. 

one,   two,   three,   four,   iive,   six,   seven,  eight,   nine,   cipher. 

The  first  nine  are  sometimes  called  digits,  and  the  cipher,  naught 
or  zero. 

29.  The  place  of  a  figure  is  its  particular  position  with 
regard  to  other  figures ;  as  in  61  (sixty-one)  counting  from 
the  right,  the  1  occupies  the  first  place  and  the  6  the  second 
place,  and  so  on  for  any  other  like  arrangement  of  figures. 

30.  The  digits  have  been  denominated  significant  figures, 
because  each  of  itself  has  a  positive  value,  always  repre- 
senting so  many  units,  or  ones,  as  .its  name  indicates.  But 
the  size  or  value  of  the  units  represented  by  a  figure  differs 
with  the  place  occupied  by  the  figure.  Thus  in  235  (two 
hundred  and  thirty-five),  each  of  the  figures,  without  regard 
to  its  place,  expresses  units,  or  ones ;  but  these  units  or  ones 
differ  in  value.     The  5  occupying  the  first  place  represents 


NOTATION.  19 

5  single  units ;  the  3  occupying  the  second  place  represents 
3  tens,  or  3  units  each  ten  times  the  size  or  value  of  a  unit  of 
the  first  place  ;  and  the  2  occupying  the  third  place  represents 
2  hundreds,  or  2  units  each  one  hundred  times  the  size  or  value 
of  a  unit  of  the  first  place  ;  the  value  of  any  figure  being  always 
increased  tenfold  by  each  removal  of  it  one  place  to  the  left. 

31.  The  cipher  becomes  significant  when  connected  with 
other  figures,  by  filling  a  place  that  otherwise  would  be  va- 
cant ;  as  in  10  (ten)  where  it  occupies  the  vacant  place  of 
units,  and  in  102  (one  hundred  and  two)  where  it  fills  the  va- 
cant place  of  tens. 

32.  The  simple  value  of  a  figure  is  the  value  its  unit 
has  when  the  figure  stands  alone ;  or,  in  a  collection,  when 
standing  in  the  right-hand  place.  Thus  2  alone,  or  in  32 
(thirty-two),  expresses  a  simple  value  of  two  single  units  or 
ones. 

33.  The  local  value  of  a  figure  is  the  value  its  unit  has 
when  the  figure  is  removed  from  the  right-hand  place,  and  de- 
pends upon  the  place  the  figure  occupies.  Thus,  in  44  (forty- 
four),  the  4  in  the  first  place  has  the  local  value  of  4  units,  and 
the  4  in  the  second  place  has  the  local  value  of  4  tens,  or  forty. 

34.  The  successive  places  occupied  by  figures  are  often 
called  orders.  Thus  a  figure  in  the  first  or  units'  place  is 
called  a  figure  of  the  first  order,  or  of  the  order  of  units  ;  a 
figure  in  the  second  place  is  a  figure  of  the  second  order,  or  of 
the  order  of  tens ;  in  the  third  place,  of  the  third  order,  or  of 
the  order  of  hundreds  ;  and  so  on,  each  figure  next  to  the  left 
belonging  to  a  distinct  order,  the  unit  of  which  is  tenfold  the 
size  or  value  of  a  unit  of  the  order  at  the  right. 

Exercises. 

1.  "Write  three  units  of  the  first  order. 

2.  Write  H\e  units  of  the  first  order. 

3.  Write  eight  units  of  the  second  order,  with  seven  of  the 
first. 

4.  Write  two  units  of  the  third  order,  with  none  of  the  sec- 
ond, and  one  of  the  first. 

5.  Write  seven  units  of  the  fourth  order,  with  two  of  the 
third,  none  of  the  second,  and  none  of  the  first. 


20  NUMERATION. 

6.  Write  one  unit  of  the  fifth  order,  with  none  of  the  four 
lower  orders. 

7.  Write  six  units  of  the  sixth  order,  five  of  the  fifth,  four  of 
the  fourth,  three  of  the  third,  one  of  the  second,  and  two  of  the 
first. 

8.  Write  one  unit  of  the  eighth  order,  with  none  of  the  seven 
lower  orders. 

9.  Write  nine  units  of  the  ninth  order,  with  six  of  each  of  the 
eight  lower  orders. 

10.  Write  two  units  of  the  twelfth  order,  with  none  of  the 
eleventh,  none  of  the  tenth,  one  of  the  ninth,  five  of  the  eighth, 
nine  of  the  seventh,  none  of  the  sixth,  none  of  the  fifth,  none  of 
the  fourth,  three  of  the  third,  none  of  the  second,  and  three  of 
the  first. 

11.  Write  three  units  of  the  fifteenth  order,  with  none  of  the 
fourteenth,  none  of  the  thirteenth,  none  of  the  twelfth,  one  of 
the  eleventh,  seven  of  the  tenth,  five  of  the  ninth,  one  of  the 
eighth,  none  of  the  seventh,  none  of  the  sixth,  five  of  the  fifth, 
three  of  the  fourth,  two  of  the  third,  two  of  the  second,  and 
seven  of  the  first. 

12.  Write  four  units  of  the  twenty-fifth  order,  with  three  of 
the  twenty-fourth,  two  of  the  twenty-third,  none  of  the  twenty- 
second,  none  of  the  twenty-first,  none  of  the  twentieth,  none  of 
the  nineteenth,  none  of  the  eighteenth,  none  of  the  seventeenth, 
five  of  the  sixteenth,  and  none  of  the  fifteen  lower  orders. 

NUMERATION. 

35.  Numeration  is  the  process  of  reading  numbers  when 
expressed  by  figures. 

36*  There  are  two  methods  of  numeration ;  the  French  and 
the  English. 

French  Numeration. 

37.  The  French  method  of  numeration  is  that  in  gen- 
eral use  on  the  continent  of  Europe  and  in  the  United 
States.  Beginning  at  the  right,  figures  occupying  more  than 
three  places  being  separated  into  as  many  groups  as  possible  of 
three  figures  each,  called  periods,  it  gives  a  disthict  name  to 
each  period. 


00 


X  o 


NUMERATION.  21 

FRENCH   NUMERATION   TABLE. 

If.  4 

§     f|      ga     ?g     id     g§     ^1 
J      o?a      cs      fi|      wg      a§      h| 

-3*>§    -3»J    -§C?.2    .fcft-    ,gM        ,gS        ^H-g    ^ 

ij03         £0-3         £     °Jj         £     °     O         £     °     «         r£     °     S         r%     °     ^        £ 

3    5    3        3    §'3         3    §    S3         S3    g^S         3    §S         3    §3         53    §,§        S3    § 'fl 

WH^  KHO1  WHO'  WHH  WHW  WHg  WHH  WHP 
7  8  9,  12  3,  4  5  6,  7  8  9,  12  3,  4  5  6,  7  8  9,  12  3. 

8th  Period.    7th  Period.    6th  Period.    5th  Period.    4th  Period.     3d  Period.     2d  Period.     1st  Period. 
Sextil-  duintil-         Quadril-       Trillions.         Billions.         Millions.     Thousands.       Units, 

lions.  lions.  lions. 

The  value  of  the  number  represented  in  the  table  is,  seven 
hundred  eighty-nine  sextillions,  one  hundred  twenty-three  quin- 
tillions,  four  hundred  fifty-six  quadrillions,  seven  hundred  eigh- 
ty-nine trillions,  one  hundred  twenty-three  billions,  four  hundred 
fifty-six  millions,  seven  hundred  eighty-nine  thousands,  one  hun- 
dred twenty-three. 

38.  The  unit  of  the  first  period,  or  right-hand  group,  is 
1  ;  of  the  second,  1  thousand ;  of  the  third,  1  million ;  of  the 
fourth,  1  billion ;  of  the  fifth,  1  trillion ;  of  the  sixth,  1  quad- 
rillion ;  of  the  seventh,  1  quintillion ;  of  the  eighth,  1  sextil- 
lion,  etc. 

The  periods  above  sextillions,  in  their  order,  are,  Septillions, 
Octillions,  Nonillions,  Decillions,  Undecillions,  Duodecillions, 
Tredecillions,  Quatuordecillions,  Quindecillions,  Sexdecillions, 
Septendecillions,  Octodecillions,  Novemdecillions,  Vigintillions, 
etc. 

Note.  —The  idea  of  number  is  the  latest  and  most  difficult  to  form.  Before 
the  mind  can  arrive  at  such  an  abstract  conception,  it  must  be  familiar  with 
that  process  of  classification  by  which  we  successively  ascend  from  individ- 
uals to  species,  from  species  to  genera,  from  genera  to  orders.  The  savage  is 
lost  in  his  attempts  at  numeration,  and  significantly  expresses  his  inability  to 
proceed  by  holding  up  his  expanded  fingers  or  pointing  to  the  hair  of  his  head. 
It  is,  indeed,  difficult  for  any  mind  to  form  an  adequate  idea  of  the  larger  num- 
bers. To  count  a  million,  at  the  rate  of  one  in  a  second,  would  require  upward 
of  twenty-three  days  of  twelve  hours  each.  A  billion  is  equal  to  a  million  a 
thousand  times  repeated,  or  a  number  so  great,  as  to  exceed  all  the  seconds  of 
time  that  would  elapse  in  thirty-two  years. 


22  NUMERATION. 

39.  To  read  numbers  represented  by  figures  according  to 
the  French  method ;  — 

Begin  at  the  right  hand,  and  point  off  the  figures  into  as  many 
periods  as  possible  of  three  places  each. 

Then,  commencing  at  the  left  hand,  read  the  figures  of  each  period, 
adding  the  name  of  each  period  excepting  that  of  units. 

Exercises. 

Read  orally,  or  write  in  words,  the  numbers  represented  by 
the  following  figures,  according  to  the  French  method :  — 


1. 

31620 

7. 

21623417 

13.     3256171894 

2. 

55216 

8. 

101315125 

14.     5162251912 

3. 

29156 

9. 

876223399 

15.     13141421211 

4. 

179213 

10. 

12345625 

16.    457863497639 

5. 

918512 

11. 

21177792 

17.    434378783434 

6. 

1219615 

12.  6665551161 

18.   7852463767445 

19. 

407000010703801 

21.     478127815016666060707 

20. 

2000700078 

01000 

22.  800800800800800800800800 

23.   127081061071081010009007007 
24.407144140070060700007101800808 

40.  To  write  numbers  in  figures  according  to  the  French 
method ;  — 

Begin  at  the  left  hand,  and  write  in  each  successive  order  the  figure 
belonging  to  it. 

If  any  intervening  order  would  otherwise  be  vacant,  fill ihe  place  by 
a  cipher. 

Exercises. 

Represent  by  figures,  and  read,  the  following  numbers,  ac- 
cording to  the  French  method :  — 

1.  Twenty-nine. 

2.  Four  hundred  and  seven. 

3.  Twenty-three  thousand  and  seven. 

4.  Five  millions  and  twenty-seven. 

5.  Seven  millions,  two  hundred  five  thousand  and  five. 

6.  Two  billions,  two  hundred  seven  millions,  six  hundred 
four  thousand  and  nine. 

7.  One  hundred  five  billions,  nine  hundred  nine  millions, 
three  hundred  eight  thousand  two  hundred  and  one. 


NUMERATION. 


23 


8.  Nine  quintillions,  eight  billions  and  forty-six. 

9.  Fifteen  quintillions,  thirty-one  millions  and  seventeen. 

10.  Five  hundred  seven  septillions,  two  hundred  three  tril- 
lions, fifty-seven  millions  and  eighteen. 

11.  Nine  nonillions,  forty-seven  trillions,  seven  billions,  two 
millions,  three  hundred  ninety-two. 

12.  Fifteen  duodecillions,  ten  trillions,  one  hundred  twenty- 
seven  billions,  twenty-six  millions,  three  hundred  twenty  thou- 
sand four  hundred  twenty-six. 

English  Numeration. 

41.  The  English  method  of  numeration  is  that  generally 
used  in  Great  Britain,  and  in  the  British  Provinces.  It  di- 
vides numbers  into  periods  of  six  figures,  and  gives  a  distinct 
name  to  each. 

ENGLISH  NUMERATION   TABLE.- 


2  a 

Si 

O  Cm 
m    O 

^    m 

85 

O 

H 


S^  a 

^H  o 

^  Cm  '^ 

m3§* 


M*      » 

O    fl 

e3 .2 

O  Cm 

a  © 

O    0 

££ 

Cm  H 

O  Cm 

.    O 

Is 


ro  ^m    cq 

Sw  g 

«M   .  O 

S^  22.2 


f«  5  r»  J3      £  jB  ^  s  rS  B 

HWHH      WHHMHpq 

OOOO  K      £      O      Q      7      1 


CQ 

h 

li 


tM  F=< 

ji 

«M  H 

OCm 

.  o 

'S  ^ 


Is 

£.2 


*S 


M     CQ 

o 


^ 
§  » 


sS 


8  J 

H  | 

"I    , 

"^    CQ    3^    »  .2 

g  a  o  g  fl^g 


39883  2,   56387  1,   351615,   12356  1 


4th  Period. 
Trillions. 


3d  Period. 
Billions. 


y 

2d  Period. 
Millions. 


1st  Period. 
Units. 


The  value  of  the  figures  in  the  above  table,  expressed  in 
words  according  to  the  English  method,  is,  Three  hundred 
ninety-eight  thousand,  eight  hundred  thirty-two  trillions;  five 
hundred  sixty-three  thousand,  eight  hundred  seventy-one  bil- 
lions ;  three  hundred  fifty-one  thousand,  six  hundred  fifteen 
millions ;  one  hundred  twenty-three  thousand  five  hundred 
sixty-one. 


24  NUMERATION. 

42.  To  read  numbers  represented  by  figures  according  to 
the  English  method ;  — 

Begin  at  the  right  hand,  and  point  off  the  figures  into  periods  of  six 
places  each. 

Then,  commencing  at  the  left  hand,  read  the  figures  of  each  period, 
adding  the  name  of  each  period  except  that  of  units. 

Exercises. 

Read  orally,  or  write  in  words,  the  numbers  represented  by 
the  following  figures,  according  to  the  English  method :  — 


1.  23457896 

2.  325487691 

3.  1678912161 


4.  98765421910311 

5.  5632411132321300012 

6.  6961771889133201443345567 


43.  To  write  numbers  in  figures  according  to  the  English 
method ;  — 

Begin  at  the  left  hand,  and  write  in  each  successive  order  the  figure 
belonging  to  it. 

If  any  intervening  order  would  otherwise  be  vacant,  fill  the  place  by 
a  cipher. 

Exercises. 

Represent  by  figures,  and  read,  the  following  numbers,  ac- 
cording to  the  English  method :  — 

1.  Thirty-two  million  three  hundred. 

2.  Seven  billion  seventeen  thousand. 

3.  Five  hundred  sixty  thousand  one  hundred  two  million, 
nine  hundred  twenty-nine  thousand  four  hundred  eleven. 

4.  One  trillion,  seven  hundred  forty-eight  thousand  nine  hun- 
dred fifty-five  billion. 


ADDITION 


44.  Addition  is  the  process  of  collecting  two  or  more 
numbers  into  one  sum.  The  result  thus  obtained  is  called 
the  amount. 

When  the  numbers  added  are  all  of  the  same  denomination, 
as  all  dollars,  all  days,  the  process  is  termed  addition  of  simple 
numbers. 


ADDITION.  25 

45.  To  add  simple  numbers. 

Ex.  1.  A  man  has  three  farms ;  the  first  contains  378  acres, 

the  second  586  acres,  and  the  third  168  acres.     How  many 

acres  are  there  in  the  three  farms.  Ans.  1132. 

operation.      Having  arranged  the  numbers  so  that  all  the  units 

Acres.      0f  the  sa«me  order  shall  stand  in  the  same  column, 

3  7  8     we  first  add  the  column  of  units  ;  thus,  8  and  6  are 

5  8  6     14,  and  8  are  22  units,  =  2  tens  and  2  units.     We 

1  6  §     write  the  two  units  under  the  column  of  units,  and 

carry  or  add  the  2  tens  to  the  column  of  tens  ;  thus, 

Ans.  113  2  2  added  to  6  make  8,  and  8  are  16,  and  7  are  23 
tens,  =  to  2  hundred  and  3  tens.  We  write  the 
3  tens  under  the  column  of  tens,  and  add  the  2  hundred  to  the 
column  of  hundreds;  thus,  2  added  to  one  make  3,  and  5  are  8,  and 
3  are  11  hundred,  =  1  thousand  and  1  hundred.  We  write  the 
1  hundred  under  the  column  of  hundreds  ;  and  there  being  no  other 
column  to  be  added,  we  set  down  the  1  thousand  in  the  thousands' 
place,  and  find  the  amount  of  the  several  numbers  to  be  1132. 

In  practice,  it  is  better  not  to  name  each  figure  added,  but  only 
the  results,  thus,  8,  14,  22  units,  —  2  tens  and  2  units,  etc. 

Rule.  —  Write  the  numbers  so  that  all  the  figures  of  the  same  order 
shall  stand  in  the  same  column. 

Add,  upward,  all  the  figures  in  the  column  of  units,  and,  if  the 
amount  be  less  than  ten,  write  it  underneath.  But  if  the  amount  be 
ten  or  more,  write  down  the  unit  figure  only,  and  add  in  the  figure 
denoting  the  ten  or  tens  ivith  the  next  column. 

Proceed  in  this  way  ivith  each  column,  until  all  are  added,  observing 
to  write  under  the  last  column  its  ivhole  amount. 

46.  First  Method  of  Proof  —  Begin  at  the  top  and  add 
the  columns  downward  in  the  same  manner  as  they  were  be- 
fore added  upward;  and  if  the  two  sums  agree,  the  work  is 
presumed  to  be  right. 

The  reason  of  this  proof  is,  that,  by  adding  downward,  the 
order  of  the  figures  is  inverted  ;  and,  therefore,  any  error  made 
in  the  first  addition  would  probably  be  detected  in  the  second. 
Note.  —  This  method  of  proof  is  generally  used  in  business. 

47.  Second  Method  of  Proof,  —  Separate  the  numbers 
to  be  added  into  two  parts,  by  drawing  a  horizontal  line  be- 
tween them.  Add  the  numbers  below  the  line,  and  set  down 
their  sum.  Then  add  this  sum  and  the  number  or  numbers 
above  the  line  together ;  and  if  their  sum  is  equal  to  the  first 
amount,  the  work  is  presumed  to  be  right. 

N    3 


26 


ADDITION. 


The  reason  of  this  proof  depends  on  the  principle,  that  the 
sum  of  all  the  parts  into  which  any  number  is  separated  is  equal 
to  the  whole.     (Art.  24,  Ax.  13.) 


Examples. 


2. 

OPERATION. 

Dollars. 

765 
381 
872 
315 


2. 

OPERATION  AND  PROOF. 

Dollars. 

765 


3. 


3. 


OPERATION.      OPERATION  AND  PROOF. 

Tons.  Tons. 

126         126 


381 
872 
315 


384 
876 
243 


384 
876 
243 


Ans.  2  3  3  3  First  ain't  2  3  3  3  Ans.  16  2  9  First  am't  16  2  9 


1568 


Ans.  2  33  3 


1503 
Ans.  16  2  9 


4. 

Barrels. 

123 
456 

789 
341 


5. 

Pounds. 

678 
901 
278 
633 


6. 

Acres. 

456 
789 
127 
815 


7. 

Cents. 

789 
987 
123 
321 


8. 

Eagles. 

456 
781 
197 
715 


9. 

Rods. 

781 
175 
564 
337 


1709     2490     2187     2220     2149     1857 
11.  12.  13.  14.  15. 

Ounces.  Inches.  Rods.  Furlongs.  Cords. 

7891  3256  6789  1234  4567 

3245  7890  1234  5678  8912 

6  7  89  1234  5678  9012  3456 

1234  5678  6543  3456  7891 

5678  7801  1234  7891  4567 


10. 

Poles. 

889 
776 
432 

876 

2973 
16. 

Feet. 

4561 
7890 
7658 
8888 
9199 


24837  25859  21478  27271  29393  38196 


17. 

Hogsheads. 

1789 
6  5  43 
2177 
8915 
6781 
4325 


18. 

Furlongs. 

6781 
1371 
8715 
6371 
1234 
7171 


19. 

Miles. 

7890 
1070 
4437 
6789 
5378 
1234 


20. 

Dollars. 

1785 
5678 
9137 
8171 
1888 
1919 


21. 

Casks. 

7895 
5678 
7186 
5176 
4321 
4127 


22. 

Pence. 

4371 
1699 
1098 
8816 
6171 
7185 


ADDITION. 


27 


23. 

Shillings. 

78956 
321  67 
41328 
45678 
13853 
71667 

28. 

Acres. 

789516 
377895 
378567 
832156 

7  8  9  5  67 

8  13  13  8 


24. 

Tons. 

12345 

87655 
34517 
65483 
79061 
20939 


25. 

Miles. 

34567 
78901 
32199 
17188 
88888 
12345 


29. 

Roods. 

451237 
813715 
679919 
787651 
6  37171 
813785 


26. 

Trees. 

76717 
77776 
67890 
71444 
47474 
16175 


30. 

Pole&r 

1234567 
8901234 
5678901 
3456789 
5432115 
7177444 


27. 

Loads. 

56789 
12345 
67819 
34567 
71888 
33197 

31. 

Yards. 

789123 

456789 
987654 
357913 
245678 
999999 


32.  What  is  the  sum  of  15  -f  26  +  18  -f  91  ?    Ans.  150. 

33.  What  is  the  sum  of  6789  -f  5832  +  4671  -f  8907  ? 

Ans.  26199. 

34.  Required  the  sum  of  76  +  48  +  59  -f  81.  Ans.  264. 

35.  Required  the  sum  of  123456  +  789012  -f  345678  + 
901234  +  567890  -f  987654  +  321032  -f  765437. 

36.  Required  the  sum  of  876543  +  789112  +  345678  + 
965887  +  445566  +  788743  +  399378  +  456789. 

37.  Required  the  sum  of  789012  -f  345678  +  901234  + 
789037  -f  891133  -f-  477666  +  557788  +  888878. 

38.  Required  the  sum  of  987654  -f  456112  +  222333  + 
456789  +  987654  -f  321178  +  123456  -f  789561. 

39.  Required  the  sum  of  678953  -f  467631  -f  117777  + 
888888  +  444444  +  667679  -f  998889  +  671236. 

40.  Required  the  sum  of  783256  +  7128  +  39  +  432815 
-|_  99  +  67851  +  125  -f  641236  +  801  +  4328. 

41.  Required  the  sum  of  12004  -f-  32  -f-  1  -f-  7836  +  100 
_(_  46  +  3  +  6176  +  32  +  91876. 

42.  Add  together  763,  4663,  37,  49763,  6178,  and  671. 

Ans.  62075. 

43.  Add  together  15,  7896,  1,  13,  106,  113,  156,  100,  2201. 


28  ADDITION. 

44.  A  butcher  sold  to  A  369  lbs.  of  beef,  to  B  1G0  lbs.,  to  C 
861  lbs.,  to  D  901  lbs.,  to  E  71  lbs.,  and  to  F  8716  lbs. ;  what 
did  they  all  receive?  Ans.  11087  lbs. 

45.  A  owes  to  one  creditor  596  dollars,  to  another  3961,  to 
another  581,  to  another  6116,  to  another  469,  to  another  506, 
to  another  69381,  and  to  another  1261.  What  does  he  owe 
them  all?  Ans.  $82871. 

46.  If  a  boy  earn  17  cents  a  day,  how  much  will  he  earn  in 
7  days?  v  Ans.  119  cts. 

47.  If  a  man's  wages  be  19  dollars  per  month,  what  are  they 
per  year  ?  Ans.  $  228. 

48.  If  a  boy  receive  a  present  every  New  Year's  day  of  1783 
dollars,  how  much  money  will  he  possess  when  he  is  21  years 
old?  Ans.  $37443. 

48.    Method  of  adding  two  or  more  columns  at  one  operation. 

Ex.  1.  A  merchant  paid  for  cloth  219  dollars,  for  flour  416 
dollars,  for  hardware  711  dollars,  and  for  rent  93  dollars.  How 
much  did  he  pay  in  all  ?  Ans.  $  1439. 

operation.  Beginnillg  with  the  number  last  written  down, 

Dollars.  ^  a jj  ^  units  an(j  teng  .   thus>  93  an(j  X  _  94^ 

219  and  10  =  104,  and  6  =  110,  and  10  =  120,  and 

416  9  =  129,  and  10  =  139.    Of  this  amount  we  write 

7  11  the  9  units  and  3  tens  under  the  columns  added  ; 

9  3  and  add   in  the  1   hundred  with   the  column  of 

■ —  hundreds  ;  thus,  1  (carried)  and  7  =  8,  and  4  = 

Ans.  14  3  9  12,  and  2  =  14  hundred,  of  which  we  write  the 

4  hundred  under  the  column  of  hundreds  and  t\\Q 
1  thousand  in  the  thousands'  place ;  and  find  the  whole  amount  to  be 
1439.  In  like  manner  may  be  added  more  than  two  columns  at  one 
operation. 

Note.  —  The  examples  that  follow  can  be  performed  as  the  above,  or  by  the 
common  method,  or  by  both. 

2.  How  many  were  the  members  of  Congress  in  1856,  there 
being  2  Senators  from  each  State,  and  Maine  sending  6  Repre- 
sentatives, New  Hampshire  3,  Massachusetts  11,  Rhode  Island 
2,  Connecticut  4,  Vermont  3,  New  York  33,  New  Jersey  5, 
Pennsylvania  25,  Delaware  1,  Maryland  6,  Virginia  13,  North 
Carolina  8,  South  Carolina  6,  Georgia  8,  Alabama  7,  Mississippi 
5,  Louisiana  4,  Tennessee  10,  Kentucky  10,  Ohio  21,  Indiana 
11,  Illinois  9,  Wisconsin  3,  Iowa  2,  Missouri  7,  Arkansas  2, 
Michigan  4,  Florida  1,  Texas  2,  California  2.  Ans.  296. 


ADDITION.  29 

3.  Water-mills  were  invented  in  the  year  555  after  Christ ; 
windmills  744  years  after  water-mills ;  pumps  126  years  after 
windmills ;  printing  15  years  after  pumps ;  watches  37  years 
after  printing  ;  the  spinning-wheel  53  years  after  watches ; 
the  steam-engine  119  years  after  the  spinning-wheel;  the  fire- 
engine  14  years  after  the  steam-engine ;  the  spinning-frame  98 
years  after  the  fire-engine ;  and  the  electro-magnetic  telegraph 
71  years  after  the  spinning-frame.  What  year  was  the  electro- 
magnetic telegraph  invented?  Ans.  1832. 

4.  During  the  American  Revolutionary  war,  the  British  lost 
at  the  battle  of  Lexington  273  men,  at  Bunker  Hill  1054  men, 
at  Long  Island  400  men,  at  White  Plains  300  men,  at  Fort 
Washington  1000  men,  at  Trenton  1020  men,  at  Princeton 
400  men,  at  Hubbardton  200  men,  at  Bennington  800  men, 
at  Brandywine  500  men,  at  Stillwater  600  men,  at  German- 
town  500  men,  at  Saratoga  400  men,  by  Burgoyne's  sur- 
render 5791  men,  at  Fort  Mercer  500  men,  at  Monmouth 
400  men,  at  Rhode  Island  260  men,  at  Brier  Creek  23  men, 
at  Stony  Point  600  men,  at  Savannah  130  men,  at  Camden 
325  men,  at  King's  Mountain  1150  men,  at  Cowpens  800 
men,  at  Guilford  Court-House  600  men,  at  Hobkirk's  Hill 
250  men,  at  Eutaw  Springs  700  men,  at  Yorktown  7000  men. 
What  was  the  entire  loss?  Ans.  25976  men. 

5.  Required  the  number  of  square  miles  in  the  following 
States,  there  being  in  Maine  35,000,  in  New  Hampshire 
8,030,  in  Vermont  8,000,  in  Massachusetts  7,250,  in  Rhode 
Island  1,200,  in  Connecticut  4,750,  in  New  York  46,000,  in 
New  Jersey  6,851,  in  Pennsylvania  47,000,  in  Ohio  39,964, 
in  Michigan  56,243,  in  Indiana  33,809,  in  Illinois  55,405, 
in  Wisconsin  53,924,  in  Iowa  50,914,  and  in  California 
188,982.  Ans.  643,322. 

6.  Required  the  number  of  square  miles  in  the  following 
States,  there  being  in  Delaware  2,120,  in  Maryland  11,000, 
in  Virginia  61,352,  in  North  Carolina  45,500,  in  South  Caro- 
lina 28,000,  in  Georgia  58,000,  in  Alabama  50,722,  in  Flor- 
ida 59,268,  in  Mississippi  47,147,  in  Tennessee  44,000,  in 
Kentucky  37,680,  in  Missouri  67,380,  in  Arkansas  52,198, 
in  Louisiana  46,431,  and  in  Texas  325,520.      Ans.  936,318. 

7.  According  to  the  census  of  1850,  Maine   had  583,169 

3* 


30  SUBTRACTION. 

inhabitants,  New  Hampshire  317,976,  Massachusetts  994,514, 
Rhode  Island  147,545,  Connecticut  370,792,  Vermont  314,120, 
New  York  3,097,394,  New  Jersey  489,555,  Pennsylvania 
2,311,786,  Delaware  91,532,  Maryland  583,034,  District  of 
Columbia  51,687,  Virginia  1,421,661,  North  Carolina  869,039, 
South  Carolina  668,507,  Georgia  906,185,  Florida  87,445, 
Alabama  771,623,  Mississippi  606,526,  Lousiana  517,762, 
Texas  212,592,  Arkansas  209,897,  Tennessee  1,002,717, 
Missouri  682,044,  Kentucky  982,405,  Ohio  1,980,329,  In- 
diana 988,416,  Illinois  851,470,  Michigan  397,654,  Wisconsin 
305,391,  Iowa  192,214,  California  92,597,  and  the  Territories 
92,298.     What  was  the  whole  number  of  inhabitants  ? 

Ans.  23,191,876. 


SUBTRACTION. 


49.  Subtraction  is  the  process  of  taking  one  number 
from  another  to  find  the  difference. 

When  the  two  numbers  are  unequal,  the  larger  is  called  the 
Minuend  and  the  less  number  the  Subtrahend ;  and  when  the 
numbers  are  equal,  either  is  the  Minuend,  and  the  other  is  the 
Subtrahend,  The  result,  or  the  number  found  by  the  subtrac- 
tion, is  called  the  Difference,  or  Remainder. 

When  the  numbers  are  of  the  same  denomination,  the  pro- 
cess is  termed  Subtraction  of  Simple  Numbers* 

50.  To  subtract  simple  numbers. 

Ex.  1.     From  935  take  673.  .  Ans.  262. 

operation.  We  first  take  the  3  units  from  the  5 
Minuend  9  3  5  units,  and  find  the  difference  to  be  2  units, 
Subtrahend  6  7  3      wnic^  we  write  under  the  figure  subtracted. 

We  then  proceed  to  take  the  7  tens  from 

Remainder  2  6  2  the  3  tens  above  it ;  but  we  here  find  a 
difficulty,  since  the  7  is  greater  than  the 
3,  and  cannot  be  subtracted  from  it.  We  therefore  add  10  tens  to  the 
3  tens,  which  makes  13  tens,  and  then  subtract  the  7  from  13,  and  6 
tens  remain,  which  we  write  below.  Then,  to  compensate  for  the 
10  tens,  equal  to  1  hundred,  added  to  the  3  tens  in  the  minuend, 
we  add  1  hundred  to  the  6  hundred  of  the  subtrahend,  which  makes 


SUBTRACTION.  31 

7  hundreds,  and  subtract  the  7  hundreds  from  9  hundreds,  and  2 
hundreds  remain.  By  adding  the  10  tens  to  the  minuend  and  the  1 
hundred  to  the  subtrahend,  the  two  numbers  being  equally  increased, 
their  difference  is  not  changed.  (Art.  24,  Ax.  9.)  The  remainder  is 
262. 

Note.  —  The  addition  of  10  to  the  minuend  is  sometimes  called  borrowing 
10,  and  the  addition  of  1  to  the  subtrahend  is  called  carrying  1. 

Rule.  —  Place  the  less  number  under  the  greater,  so  that  units  of 
the  same  order  shall  stand  in  the  same  column. 

Commencing  at  the  right  hand,  subtract  each  figure  of  the  subtrahend 
from  the  figure  above  it. 

If  any  figure  of  the  subtrahend  is  larger  than  the  figure  above  it  in 
the  minuend,  add  10  to  that  figure  of  the  minuend  before  subtracting, 
and  then  add  1  to  the  next  figure  of  the  subtrahend. 

51  •  First  Method  of  Proof  . — Add  the  remainder  and  the 
subtrahend  together,  and  their  sum  will  be  equal  to  the  min- 
uend, if  the  work  is  right. 

This  method  of  proof  depends  on  the  principle,  that  the 
greater  of  any  two  numbers  is  equal  to  the  less  added  to  the 
difference  between  them. 

52.  Second  Method  of  Proof.  —  Subtract  the  remainder  or 
difference  from  the  minuend,  and  the  result  will  be  like  the 
subtrahend,  if  the  work  is  right. 

This  method  of  proof  depends  on  the  principle,  that  the 
smaller  of  any  two  numbers  is  equal  to  the  remainder  obtained 
by  subtracting  their  difference  from  the  greater. 


Examples. 

M* 

2. 

2. 

3. 

3. 

OPERATION.     OPERATION  AND   PROOF.      OPERATION.     OPERATION  AND  PROOF. 

Minuend       4  6  9 

469 

788 

788 

Subtrahend  18  3 

183 

369 

369 

Remainder  2  8  6 

286 

419 

419 

Min. 

469 

Sub. 

369 

4. 

5. 

6. 

7. 

Miles. 

Gallons. 

Minutes. 

Pecks. 

From          7  6  5  4 

7116 

6178 

4567 

Take           19  7  8 

1997 

17  69 

1978 

32 


SUBTRACTION. 


8. 
Barrels. 

From  7  6  5  116 
Take  7  16  6  6  9 


12. 

Hogsheads. 

From  6110  0  0 
Take    19  9999 


16. 

Hoods. 

From  10  0  2  0  0 
Take      9  8  7  6  1 


9. 

Degrees. 

56789 
10091 

13. 

Bushels. 

617853 
190909 

17. 

Acres. 

511799 
419109 


10. 

Furlongs. 

56781 
39109 

14. 

Yards. 

7111111 
909009 

18. 

Poles. 

610000 
166666 


From 
Take 


From 
Take 


From 
Take 


20. 

Dollars. 

10000000 
9099019 


21. 

Eagles. 

99999999 
1000919 


11. 

Tons. 

71678 
18819 

15. 

Pounds. 

999000 
199919 

19. 

Cords. 

789111 
171670 

22. 

Guineas. 

888888 
99999 


23. 

Seconds. 

100200300400500 
90807060504039 


25. 

Months. 

61567101 
91678 


26. 

Days. 

1000000 

1 


24.    w 

Hours. 

600700800900 
19181891718  5 

27. 

"Weeks. 

10000000 
9999999 


28.  What  is  the  value  of  6767851  —  81715  ? 

29.  What  is  the  value  of  761619161  —  916781  ? 

30.  What  is  the  value  of  31671675  —  361784? 

31.  What  is  the  value  of  16781321  —  100716? 

32.  What  is  the  value  of  1002007000  —  5971621  ? 

33.  Sir  Isaac  Newton  was  born  in  the  year  1642,  and 
he  died  in  1727 ;  how  old  was  he  at  the  time  of  his  de- 
cease? Ans.  85  years. 

34.  Gunpowder  was  invented  in  the  year  1330 ;  how  long 
was  this  before  the  invention  of  printing,  which  was  in 
1440?  Ans.  110  years. 


SUBTRACTION.  33 

35.  The  mariner's  compass  was  invented  in  Europe  in  the 
year  1302 ;  how  long  was  this  before  the  discovery  of  America 
by  Columbus,  which  happened  in  1492  ?         Ans.  190  years. 

36.  What  number  is  that,  to  which  if  6956  be  added,  the 
sum  will  be  one  million?  Ans.  993044. 

37.  A  man  bought  an  estate  for  seventeen  thousand  five 
hundred  and  sixty-five  dollars,  and  sold  it  for  twenty-nine 
thousand  three  hundred  and  seventy-five  dollars.  Did  he  gain 
or  lose,  and  how  much?  Ans.  Gained  $  11810. 

38.  Bonaparte  was  declared  emperor  in  1804 ;  how  many 
years  since  ? 

39.  The  union  of  the  government  of  England  and  Scotland 
was  in  the  year  1603;  how  long  was  it  from  this  period  to 
1776,  the  time  of  the  declaration  of  the  independence  of  the 
United  States?  Ans.  173  years. 

40.  Jerusalem  was  taken  and  destroyed  by  Titus  in  the  year 
70 ;  how  long  was  it  from  this  period  to  the  time  of  the  first 
Crusade,  which  was  in  the  year  1096?         Ans.  1026  years. 

41.  From  the  Creation  to  the  Deluge  was  1656  years; 
thence  to  the  founding  of  Rome  1595  years;  thence  to  the 
death  of  Charlemagne,  which  took  place  814  years  after 
Christ,  1567  years.  In  what  year  of  the  world  was  Christ 
born?  Ans.  4004. 

42.  A  gentleman  83  years  old  has  two  sons ;  the  age  of  the 
older  son  added  to  his  makes  -128  years,  and  the  age  of  the 
younger  son  is  equal  to  the  difference  between  the  age  of  the 
father  and  that  of  the  older  son.  How  old  is  each  of  his 
sons  ?         Ans.  The  older,  45  years  ;  the  younger,  38  years. 

43.  During  the  year  1810,  there  were  manufactured  in  the 
United  States  one  hundred  and  forty-six  thousand  nine  hun- 
dred and  seventy-four  yards  of  cotton  cloth;  and  during  the 
year  1855,  five  hundred  and  twenty  million  yards.  What  was 
the  increase?  Ans.  519,853,026  yards. 

53.  Method  of  subtracting,  when  there  are  two  or  more 
subtrahends. 

Ex.  1.  From  a  pile  of  wheat  containing  657  btrshels,  A  is 
to  have  141  bushels,  B  244  bushels,  C  134  bushels,  and  D  the 
remainder.     How  many  bushels  is  D  to  have  ?         Ans.  138. 


34  SUBTRACTION. 

FIRST  OPERATION.  SECOND  OPERATION.        Ill     the      first     OpOTcl- 

Busheis.  Bushels,  tion,   the  several  sub- 

Minuend       6  5  7  Minuend       6  5  7  trahends  are  added  for 

"TTT  s  i  a  -i  a  single  subtrahend  to 

141  \  1 4  ]  be    taken     from    the 

2  4  4  -^244  minuend.    In  the  sec- 

13  4  ^13  4  ond,    the    subtrahends 

Subtrahend  ilo       Remainder  188     Z%%%£$. 

Remainder   13  8  at}oni  ^us :  beginning 

with  units,  4  and  4  and 
1  =  9,  which  from  17  units  leaves  8  units;  passing  to  tens,  1 
(carried)  and  3  and  4  and  4  =  12  tens;  reserving  the  left-hand 
figure  to  add  in  with  the  figures  of  the  subtrahends  in  the  next 
column,  the  right>hand  figure,  2  tens,  which  we  subtract  from  the 
5  tens  of  the  minuend,  and  have  left  3  tens ;  and,  passing  to  hun- 
dreds, we  add  in  the  left-hand  figure  1 ,  reserved  from  the  1 2  tens, 
which  with  the  other  figures  1  and  2  and  1  =  5  hundreds,  which,  taken 
from  6  hundreds,  leaves  1  hundred;  and  138  is  the  answer  sought. 

2.  John  Drew  has  a  yearly  income  of  2,500  dollars;  his 
family  expenses  are  1,300  dollars,  his  expenditures  in  improv- 
ing his  estate  450  dollars,  and  his  contributions  to  several 
worthy  objects  225  dollars.  What  remains  to  lay  up  or  in- 
vest? 

3.  A  speculator  bought  four  village  lots ;  for  the  first  he  paid 
620  dollars;  for  the  second,  416  dollars;  for  the  third,  350 
dollars ;  for  the  fourth,  225  dollars ;  and  sold  the  whole  for 
2,000  dollars.     What  did  he  gain  ? 

4.  Daniel  White,  dying,  left  property  to  the  amount  of 
27,563  dollars,  of  which  his  wife  received  9,188  dollars,  each 
of  his  two  daughters,  4,594  dollars,  and  his  only  son  the  bal- 
ance.    What  did  his  son  receive?  Ans.   9,187  dollars. 

5.  The  United  States  contain  2,983,153  square  miles,  of 
S  which  the  Atlantic  slope  includes   967,576,  the  Pacific  slope 

778,266,   and   the   Mississippi  Valley  the   remainder.     How 
many  square  miles  does  the  Mississippi  Valley  contain  ? 

Ans.  1,237,311. 

6.  The  British  North  American  Provinces  contain  3,125,401 
square  miles;  of  which  147,832  square  miles  belong  to  Canada 
West ;  201,989  to  Canada  East ;  27,700  to  New  Brunswick  ; 
18,746  to  Nova  Scotia;  2,134  to  Prince  Edward's  Island; 
57,000  to  Newfoundland ;  170,000  to  Labrador ;  and  the  re- 


MULTIPLICATION.  35 

mainder  to  the  Hudson's  Bay  Territory.     What  number  of 
square  miles  belong  to  the  Hudson's  Bay  Territory? 

Ans.  2,500,000. 

7.  James  Howe  has  property  to  the  amount  of  63,450  dollars, 
and  owes  in  all  three  debts ;  one  of  1000  dollars,  another  of  350 
dollars,  and  another  of  12,468  dollars.  How  much  has  he  after 
paying  his  debts  ? 

8.  The  entire  coinage  of  the  mint  of  the  United  States,  in- 
cluding the  coinage  of  its  branches,  from  1792  to  1856,  amounted 
in  value  to  $498,197,382,  of  which  $396,895,574  was  gold, 
$  100,729,602  silver,  and  the  remainder  of  the  amount  cop- 
per.    What  was  the  value  of  the  copper  coinage  ? 

Ans.  $  572,206. 


MULTIPLICATION. 

54.  Multiplication  is  the  process  of  taking  one  number 
as  many  times  as  there  are  units  in  another  number. 

In  multiplication  three  terms  are  employed,  called  the  Multi- 
plicand, the  Multiplier,  and  the  Product. 

The  multiplicand  is  the  number  to  be  multiplied  or  taken. 

The  midtiplier  is  the  number  by  which  we  multiply,  and 
denotes  the  number  of  times  the  multiplicand  is  to  be  taken. 

The  product  is  the  result,  or  number  produced  by  the  multi- 
plication. 

The  multiplicand  and  multiplier  together  are  called  factors, 
from  the  product  being  made  or  produced  by  them. 

When  the  multiplicand  consists  of  a  single  denomination,  the 
process  is  termed  multiplication  of  simple  numbers. 

In  the  following  table,  the  invention  of  Pythagoras,  may  be 
found  all  the  elementary  products  necessary  in  performing  any 
operation  in  multiplication,  since  the  multiplication  of  numbers, 
however  large,  depends  upon  the  product  of  one  digit  by  an- 
other. The  products,  therefore,  of  each  digit  by  any  other, 
should  be  thoroughly  committed  to  memory.  Considerable 
more  of  the  table,  even,  may  be  memorized  with  fully  com- 
pensating results. 


86 


MULTIPLICATION. 


MULTIPLICATION  TABLE. 


m 

CM 

o 

P 
P 

71 

o 

17 

— 

P 
P 

71 

m 

71 
71 

o 
in 

71 

K3 
5 

P 

P 

CO 

17 
71 

eg 

P 
■  7 
CO 

O 

CO 

o    o 

O     CN 

P   m 

O    >n    o    m   o 
o    cn    m   b-    o 
m   m    in    in   co 

m 

CN 

CO 

01 

00 

xf 

71 

CO 

OS 

P 
71 

oo 
to 

71 

2 

to 

1— 1 
71 

? 

71 

to 

71 

CO 
X. 

71 

71 

C7 

CO 

CO 
CO 

O 

CO' 
CO' 

Xj 
CO 

XJ 

c 

CM     CO 

CO     o 

O     t*     00     CN     CO 

oo    o    cn    in    r^ 
rf    o    m    m    in 

o 

o 

CO 

CO 

CO 

OS 

CO 

OS 

m 

go 

C7 

to 

1- 

P 
71 

O 
CO 
71 

CO 
i.7 
71 

CO 
I- 
71 

OS 
OS 
71 

71 
71 

C7 

00 

CO 
C7 

■— 1 

OS 
CO 

i-h     CO 

O     CO 
CO     00 

CO     OS     CN 

O    cm    m 

m   m   m 

m 
m 

CN 

to 

CO 

GO 
00 

o 

71 

•<* 

.7 

00 

OS 

CN 

71 
t 
71 

71 

to 

X 

71 

00 

S 

CO 

C7 

71 

i7 

CO 

C7 

§  2 

CO     "* 

O     CM 

^     CO 

00 

CO     00 
O     CN 

m    in 

o 
m 
m 

CM 

71 

w 

(O 

3 

m 

o 

to 

71 

GO 

OS 

2 

P 
71 

CO 
71 

71 
»-7 
71 

C7 
I- 
71 

OS 

71 

.7 

to 

CO 
CO 

1-7 

CO 

00     OS 
t^.    OS 

CO     CO 

O        ^-> 

CN     t* 

71 

CO 

CO     -T 
00    o 
■<*    in 

m 

CN 

m 

O 
CM 

P 

- 
-o 

do 

71 

-+ 

t 

71 

71 

71 

71 

71 

p 

C7 

CO 

3 

CO 

CO     CO 

O     CN 

c 

P 
to 

C 
00 

o 
m 

2 

x 

o 

co 
t 

9) 

-r 

CO 

77 

71 

.7 

- 

I- 

O 
OS 

OS 
P 
71 

X 

71 
71 

I- 

^! 

71 

to 
to 

71 

i7 
Xj 
71 

5 

C7 

CO 
71 

C7 

CN  —h 

-*     CO 

co    co 

O    OS 
00     OS 

co   co 

X 

<-* 

t^.     CO 

co    m 

m 

00 

m 

I- 

OS 

O 

71 

to 

71 

eg 

71 

71 

1- 
71 

71 

C7 

CN  ,««* 

co    co 

CO    t^ 

co    co 

OS 
C7 

rf 

^ 

71 

eg 

m 

t-^ 

1.7 

00 

OP 

is 

71 

a 

at 
— 

CO 

eg 

eg 

i7 

O 
1- 

X 

g 

71 

71 
71 

X 

C7 
71 

i7 
71 

71 
t- 

71 

OS 

tc 

71 

CO     CO 
O     CN 

CO     CO 

o   «^ 

t*    in 

CO     CO 

C7 

at 

C7 

00 

~ 

~r 

m 

CN 

co 

71 

CO 

B0 

to 

n 

at 

71 

-r 

I- 

OS 

g 

71 

71 
71 

-7 
71 

i7 
71 

I- 

71 

00     O 
CN     CO 

CM     CO 

co    co 

i.7 
C7 

CO 
C7- 

00 

CO 

O 

m 

p 

cc 

<* 

Q 

-o 

n 
t- 

OS 

■  7 

o 

VI 

.7 
C7 

.7 

1.7 

to 

C7 
X 

.7 

o 

71 

•  7 
71 
71 

a 

71 

1.7 
17 
71 

o   »n 

t^     00 
CN     CN 

8  2 

CO     CO 

C7 

.7 

3 

P 
CO 
CO 

m 

CO 

-t 

00 

71 

7i 

m 

t^ 

X 

- 

71 

-f 

i7 

-X 

X 

o. 

71 

71 

71 

C7 
71 

m    to 

CM      CN 

00     OS 
CN     CN 

g 

C7 

71 
CO 

CO 

3 

m 

CO 

CO 

CO 

71 

eg 

oi 

■a 

17 

"O 

X 

t>. 

01 

— 

- 

c 

CO 

•-o 

»7 

o. 

to 

71 
X 

•  7 

OS 

as 

71 

71 
71 

CO     't 
CN     CN 

O     CO 

CO      t- 

CN     CN 

CO 
X 
71 

OS 
O 
71 

2 

CO 

m 

CN 

CO 

CN 

71 

to 

C7 

<* 

CO 

i  - 

X 

at 

c 

M 

:r 

4 

.7 

to 

X 

OS 

fl 

s  a 

—      i7 
CN     CN 

CO 
71 

71 

00 

00 
71 

s 

CO 

i— i 

I— < 

71 
71 

3 

•  ■7 

10 

1- 

x 

00 

35 

OS 

P 

— 

71 

71 

C7 

5 

i7 

17 

■O 

CO 
1- 

I- 
X 

oo   o 

OS    o 
—     CN 

O      l-N 

71     CO 
CN     CN 

9 

71 

CO 
1.7 
71 

CO 

71 

m 

CN 

o 

as 

O 

eg 

* 

i.7 

to 

t^. 

H 

OS 

1 

71 

C7 

-7 

'7 

CO 

j: 

CO     OS 

§  2 

CN     CN 

71 

V, 

P 
71 

m 

CN 

OS 

00 

71 

to 

cr 

-t 

in 

C7 
CO 

71 
t> 

X 

o. 

OS 
O- 

X, 

I  - 

CO 

71 

17 
C7 

C7 
1.7 

CN     F-. 
co    t^ 

O    OS 
CO     CO 

00 

OS 

o 

71 

CO 

01 

m 

CN 
CN 

00 

CO 

ct 

-r 

«f 

i-7 

•x> 

I' 

x 

X 

OS 

r 

— 

71 

00 
71 

C7 

^  !m 

CO     CO 

1- 

00 

z: 

O 
CN 

I-*. 

-t 

71 

x 

71 

>.7 

eg 

71 

rr 

oT 

«o~ 

i.7 

c7 

CO 

1- 

X 

OS 

X 

o. 

i7 

3 

71 

oT 

CO     CO 
CN     CO 

I 

to 

00 
CO 

in" 

CO 

71 

30 

eg 

eg 

<* 

* 

L-7 

to 

CO 

l^ 

i- 

X 

81 

" 

6 

O     "-     CN     CN 

CO 

C7 

— 

o" 
m 

m 

2 

<—< 

P 
71 

>7 
71 

>7 
C7 

i7 

u7 

17 

CO 

■  7 

CO 

o" 

I- 

i7 
I- 

a 
So 

o    o    io 

00     OS  '  OS 

o    m 

O    P 

o 

in* 

D 

71 

in 

<tf 

X 

— < 

,N 

ON 

71 

CO 

c 

" 

t 

« 

l7 

« 

CO 

co    co    r-   t^    oo    oo 

00 

OS 

91 

O 

03 

* 

os 

71 

- 

00 

71 

7) 

71 

P 
CO 

eg 

CO 

to7 

CO 

oT 

w 

71 

i-7 
f 

x^ 

-r 

h    •*   n   o   n 

o    o    O    CO    CO 

15 
CO 

03 

t7 

in" 

oT 

^ 

to 

- 

p— i 

71 

71 

(M 

71 

71 

CC 

CO     CO 

CO     00     O     CN 

co    co    ■**  j  ^ 

^ 

^ 

-t 

in 

r— i 

o7 

CO 

«* 

in 

to 

r^ 

X 

OS 

P 

2" 

71 

7^ 

-7 

= 

CO      »>• 

ob~  ~cb   o"  — 

71 
71 

CO 
71 

71 

Tn" 

CN 

For  example,  suppose  we  wish  to  find  the  product  of  7  by  5  ; 
we  look  for  7  at  the  top  of  the  table,  and  for  5  at  the  left  hand, 
and  where  the  lines  intersect  is  35,  the  number  sought ;  or,  we 
may  look  for  the  7  at  the  left  hand,  and  the  5  at  the  top,  and 
find,  where  the  lines  intersect,  the  same  result. 


MULTIPLICATION.  37 

55.  The  repeated  addition  of  a  number  to  itself  is  equiva- 
lent to  a  multiplication  of  that  number.  Thus,  7  —J—  7  — j—  7  — J- 
7  is  equivalent  to  7  X  4,  the  sum  of  the  former  and  the  product 
of  the  latter  being  the  same.  Hence  multiplication  has  some- 
times been  called  a  concise  method  of  addition, 

56.  The  product  must  be  of  the  same  kind  or  denomination 
as  the  multiplicand,  since  the  taking  of  a  quantity  any  number 
of  times  does  not  alter  its  nature.  Thus :  5,  an  abstract  number, 
X  3  =  15,  an  abstract  number  ;  and  9  yards  X  7  =  63  yards. 

57.  The  multiplier  must  always  be  considered  as  an  abstract 
number.  Thus,  in  finding  the  cost  of  4  boohs  at  9  dollars  each, 
we  cannot  multiply  books  and  dollars  together,  which  would  be 
absurd,  but  we  can,  by  regarding  the  4  as  an  abstract  number, 
take  the  9  dollars,  or  cost  of  1  book  4  times,  and  the  product, 
36  dollars,  will  be  the  result  required. 

58.  The  product  of  two  factors  will  be  the  same,  whichever 
is  taken  as  the  multiplier.  Thus,  8x6=  6x8  =  48;  and 
the  cost  of  5  hats  at  2  dollars  each  gives  the  same  product  as 
2  hats  at  5  dollars  each.  Also,  the  product  of  any  number  of 
factors  is  the  same,  in  whatever  order  they  are  multiplied. 
Thus,  2X3X5  =  3X5X2  =  5X2X3  =  30. 

59.  A  Composite  number  is  a  number  produced  by  multi- 
plying together  two  or  more  numbers  greater  than  1.  Thus, 
10  is  a  composite  number,  since  it  is  the  product  of  2  X  5  ;  and 
18  is  a  composite  number,  since  it  is  the  product  of  2  X  3  X  3. 

60.  To  multiply  simple  numbers. 

Ex.  1.     Let  it  be  required  to  multiply  1538  by  9. 

Ans.  13842. 
operation.         Having  written  the  multiplier,  9,  un- 
Multiplicand    15  3  8       der  the  unit  figure  of  the  multiplicand, 
Multiplier  9       we  multiply  the  8  units  by  the  9,  ob- 

-r>     j  taining  72  units  =  7  tens  and  2  units. 

Product  13  8  4  2       We  wr;te  <jown  the  2  units  in  the  units' 

place,  and  reserve  the  7  tens  to  add  to 
the  product  of  the  tens.  We  then  multiply  the  3  tens  by  9,  obtaining 
27  tens,  and,  adding  the  7  tens  which  were  reserved,  we  have  34  tens 
=  3  hundreds  and  4  tens.  We  write  down  the  4  tens  in  the  tens' 
place,  and  reserve  the  3  hundreds  to  add  to  the  product  of  the.  hun- 
dreds. We  next  multiply  the  5  hundreds  by  9,  obtaining  45  hundreds, 
and,  adding  the  3  hundreds  which  were  reserved,  we  have  48  hun- 
N    4 


38  MULTIPLICATION. 

dreds  =  4  thousands  and  8  hundreds.  We  write  down  the  8 
hundreds  in  the  hundreds'  place,  and  reserve  the  4  thousands  to  add 
to  the  product  of  the  thousands.  By  multiplying  the  1  thousand  by  9 
we  obtain  9  thousands,  and,  adding  the  4  thousands  reserved,  we 
have  13  thousands,  which  we  write  down  in  full ;  —  and  the  product 
is  13842. 

2.  Let  it  be  required  to  multiply  2156  by  423. 

Ans.  911988. 
operation.  ln  this  example  the  multiplicand  is 

Multiplicand      2  15  6        to  be  taken  423  times  =  3  units  times 
Multiplier  4  2  3       -\-  2  tens  times  -(-  4  hundreds  times. 

3  units  times  2156  =  6468  units;   2 

Partial     )  tens  times  2156  =  4312  tens>  an(1  4 

Products  1       4  3  12  hundreds  times  2156  =  86 24  hundreds; 

(8624  the  sum  of  which  partial  products  = 

-p     ,  0  ..  -  a  Q  Q        911988,  or  the  total  product  required. 

1  roduct       J  1  1  J  8  8        In  the  operation  the  right-hand  figure 

of  each  partial  product  is  written  di- 
rectly under  its  multiplier,  that  units  of  the  same  order  may  stand  in 
the  same  column,  for  convenience  in  adding. 

Rule.  —  Write  the  multiplier  under  the  multiplicand,  arranging 
units  under  units,  tens  under  tens,  jrc, 

Multiply  each  figure  of  the  multiplicand  by  each  figure  of  the 
midtiplier,  beginning  with  the  right-hand  figure,  writing  the  right-hand 

figure  of  each  product  underneath,  and  adding  the  left-hand  figure  or 
figures,  if  any,  to  the  next  succeeding  product. 

If  the  multiplier  consists  of  more  than  one  figure,  the  right-hand 
figure  of  each  partial  product  must  be  placed  directly  under  the  figure 
of  the  multiplier  that  produces  it.  The  sum  of  the  partial  products  will 
be  the  whole  product  required. 

Note. — When  there  are  ciphers  between  the  significant  figures  of  the 
multiplier,  pass  over  them  in  the  operation,  and  multiply  by  the  significant 
figures  only,  remembering  to  set  the  first  figure  of  the  product  directly  under 
the  figure  of  the  multiplier  that  produces  it. 

61 .  First  Method  of  Proof.  —  Multiply  the  multiplier  by  the 
multiplicand,  and,  if  the  result  is  like  the  first  product,  the  work 
is  supposed  to  be  right.     (Art.  58.) 

62.  Second  Method  of  Proof  \ — Divide  the  product  by  the 
multiplier,  and,  if  the  work  is  right,  the  quotient  will  be  like 
the  multiplicand. 

Note.  —  This  is  the  common  mode  of  proof  in  business ;  but,  as  it  antici- 
pates the  principles  of  division,  it  cannot  be  employed  without  a  previous 
knowledge  of  that  process. 

63.  Third  Method  of  Proof.  —  Begin  at  the  left  hand  of  the 


MULTIPLICATION.  39 

multiplicand,  and  add  together  its  successive  figures  toward  the 
right,  till  the  sum  obtained  equals  or  exceeds  the  number  nine. 
If  it  equals  it,  drop  the  nine,  and  begin  to  add  again  at  this 
point,  and  proceed  till  you  obtain  a  sum  equal  to,  or  greater 
than,  nine.  If  it  exceeds  nine,  drop  the  nine  as  before,  and 
carry  the  excess  to  the  next  figure,  and  then  continue  the  ad- 
dition as  before.  Proceed  in  this  way,  till  you  have  added  all 
the  figures  in  the  multiplicand  and  rejected  all  the  nines  con- 
tained in  it,  and  write  the  final  excess  at  the  right  hand  of  the 
multiplicand. 

Proceed  in  the  same  manner  with  the  multiplier,  and  write 
the  final  excess  under  that  of  the  multiplicand.  Multiply  these 
excesses  together,  and  place  the  excess  of  nines  in  their  product 
at  the  right. 

Then  proceed  to  find  the  excess  of  nines  in  the  product 
obtained  by  the  original  operation ;  and,  if  the  work  is  right, 
the  excess  thus  found  will  be  equal  to  the  excess  contained  in 
the  product  of  the  above  excesses  of  the  multiplicand  and  mul- 
tiplier. 

Note.  —  This  method  of  proof,  though  perhaps  sufficiently  sure  for  common 
purposes,  is  not  always  a  test  of  the  correctness  of  an  operation.  If  two  or 
more  figures  in  the  work  should  be  transposed,  or  the  value  of  one  figure  be 
just  as  much  too  great  as  another  is  too  small,  or  if  a  nine  be  set  down  in  the 
place  of  a  cipher,  or  the  contrary,  the  excess  of  nines  will  be  the  same,  and 
still  the  work  may  not  be  correct.  Such  a  balance  of  errors  will  not,  however, 
be  likely  to  occur. 

Examples. 
3.  Multiply  7325  by  3612.  Ans.  26457900. 

OPERATION.  PROOF  BY  MULTIPLICATION. 

Multiplicand  7  3  2  5  Multiplicand         3  6  12 

Multiplier  3  612  Multiplier  7  3  2  5 

14650  18  0  6  0 

7325  7224 

43950  10836 

21975  25284 


Product   26457900      Product  26457900 

4.  Kequired  the  product  of  82967  by  652. 

Ans.  54094484. 


40 


MULTIPLICATION. 


Multiplicand 
Multiplier 


Product 


OPERATION. 

82967 
652 

165934 
414835 
497802 

54094484 


PROOF  BY  THE  NINES. 

5  excess. 
4  excess. 


201 


2  excess. 


2  excess. 


5. 
789123 
4 

9. 
678954 
24 


6. 
1234567 
5 

10. 
616783 
36 


989898 
2 

11. 

789563 
57 


8. 
3  7  8  9  5  8  8 
8 

12. 
789567 
98 


13.  . 
892001 
329 


14. 
230442 
701 


15. 
425016 
645 


16. 
5061029 
3408 


17.  What  will  365  acres  of  land  cost  at   73  dollars  per 
acre?  Ans.  $26645. 

18.  What  will  97  tons  of  iron  cost  at  57  dollars  a  ton? 

Ans.  $5529. 

19.  What  will  397  yards  of  cloth   cost  at  7   dollars  per 
yard?  Ans.  $2779. 

20.  What  will  569  hogsheads  of  molasses  cost  at  37  dollars 
per  hogshead?  Ans.  $21053. 

21.  If  a  man  travel  37  miles  in  one  day,  how  far  will  he 
travel  in  365  days?  Ans.  13505  miles. 

22.  If  a  vessel  sails  169  miles  in  one  day,  how  far  will  she 
sail  in  144  days? 

23.  What  will  698  barrels  of  flour  cost  at  7  dollars  a  bar- 
rel? 

24.  What  will  376  lbs.  of  sugar  cost  at  13  cents  a  pound  ? 

Ans.  4888  cts. 

25.  What  will  97  lbs.  of  tea  cost  at  93  cents  a  pound? 

Ans.  9021  cts. 


MULTIPLICATION.  41 

26.  If  a  regiment  of  soldiers  consists  of  1128  men,  how- 
many  men  are  there  in  an  army  of  53  regiments  ? 

Ans.  59784. 

27.  What  is  the  product  of  75432  X  47.     Ans.  3545304. 

28.  What  is  the  product  of  76785316  X  7615. 

Ans.  584720181340. 

29.  What  is  the  product  of  67853  X  8765. 

Ans.  594731545. 

30.  What  is  the  product  of  3812345  X  31243. 

Ans.  119109094835. 

31.  What  is  the  product  of  40670007  X  10002. 

Ans.  406781410014. 

32.  What  is  the  product  of  31235678  X  10203. 

Ans.  318697622634. 

33.  What  is  the  product  of  76786321  X  3007. 

Ans.  230896467247. 

34.  What  is  the  product  of  6176777  X  22222. 

Ans.  137260338494. 

35.  What  is  the  product  of  7060504  X  30204. 

Ans.  213255462816. 

36.  Multiply  88888  by  4444.  Ans.  395018272. 

37.  Multiply  7008005  by  10008.  Ans.  70136114040. 

38.  Multiply  987648  by  481007.        Ans.  475065601536. 

39.  Multiply  101010101  by  202020202. 

Ans.  204060808060402. 

40.  Multiply  304050607  by  3011101. 

Ans.  915527086788307. 

41.  Multiply  908007004  by  500123. 

Ans.  454115186861492. 

42.  Multiply  2003007001  by  6007023. 

Ans.  12032109124168023. 

43.  Multiply  9000006  by  9000006. 

Ans.  81000108000036. 

44.  A  full-grown  elm  will,  it  is  computed,  yearly,  on  an 
average,  produce  three  hundred  twenty-nine  thousand  three 
hundred  and  seventy-five  seeds.  How  many  seeds  will  three 
such  trees  produce  in  fifty-three  years.  Ans.  52370625.  • 

45.  John  Alden  can  plant  3  plats  of  corn,  containing  11 
rows  of  67  hills  each,  in  1  day,  and  Loring  Blanchard  can 

4* 


42  MULTIPLICATION. 

plant  twice  as  much  in  the  same  time.     How  many  hills  can 
Blanchard  plant  in  a  month  of  26  working  days  ?  Ans.  114972. 

46.  If  the  multiplicand  be  three  hundred  and  seventy-five 
millions  two  hundred  and  ninety-six  thousand  three  hundred 
and  twenty-one,  and  the  multiplier  seventy-nine  thousand  and 
twenty-four,  what  will  be  the  product  ? 

Ans.  29657416470704. 

64.     When  the  multiplier  is  a  composite  number. 

Ex.  1.     What  cost  35  acres  of  land  at  316  dollars  an  acre? 

Ans.  11060  dollars. 

operation.  The  factors  of  35  are 

3  16  dollars,  cost  of  1  acre.         7  and  5.     Now    if  we 

7  multiply  the  price  of  one 

921  2  dollirs  cost  of  7  acres         a^re  by  7'  we  3et  the  cost 
1 1 1  I  dollars,  cost  01  /  acres.       of  7  ^reg.  and?  the^  hy 

, ~  multiplying  the  cost  of  7 

110  6  0  dollars,  cost  of  35  acres.     ?cres.  J>y  &e  fact°r  5>  {t 

is  evident,  we  obtain  the 

cost  of  5  times  7  acres,  or  35  acres.    Hence,  when  the  multiplier  is  a 

composite  number,  we  may 

Multiply  the  multiplicand  by  one  of  the  factors  of  the  multiplier,  and 
the  product  thus  obtained  multiply  by  another,  and  so  on  until  each  of 
the  factors  has  been  used  as  a  multiplier :  and  the  last  product  will  be 
the  one  sought. 

Examples. 

2.  Multiply  3121  by  81,  using  its  factors. 

3.  What  will  63  horses  cost  at  175  dollars  each? 

4.  A  certain  house  contains  87  windows,  and  each  win- 
dow has  32  panes  of  glass.  How  many  panes  in  the  whole 
house?  Ans.  2784. 

5.  What  is  the  product  of  47134987  by  56? 

Ans.  2639559272. 

6.  If  a  garrison  consume  6231  pounds  of  bread  in  1  day, 
how  many  pounds  will  the  same  consume  in  144  days  ? 

Ans.  897264  pounds. 

C5.  When  there  are  ciphers  on  the  right  in  the  multiplier 
or  multiplicand,  or  both. 


MULTIPLICATION. 


43 


OPERATION. 


Ex.  1.    In  1  yard  there  are  36  inches;  how  many  inches  in 
10  yards?     In  100  yards?  Ans.  360,  3600. 

We  annex  one  cipher  to  the 
multiplicand  to  multiply  it  by 
10,  and  two  ciphers  to  multiply 
it  by  100 ;  since  annexing  one 
cipher  removes  each  figure  of 
the  multiplicand  one  place  to 
the  left,  and  thus  increases   it 


Multiplicand 
Multiplier 

Product 
Or  thus : 


36 
10 

360 
3  6  0, 


36 
100 

3T00 
3  600. 


10  times ;  annexing  two  ciphers  removes  each  figure  two  places  to  the 
left,  and  increases  it  100  times  ;  and  so  on,  each  additional  cipher  hav- 
ing the  effect  to  increase  its  value  10  times  (Art.  30). 

2.  What  will  700  bales  of  cotton  cost  at  40  dollars  per  bale  ? 

Ans.  28000  dollars. 


Multiplicand 
Multiplier 

Product 


OPERATION. 

700 
40 


28000 


The  multiplicand  we  resolve  into  the 
factors  7  and  100,  and  the  multiplier 
into  the  factors  4  and  10.  Now,  it  is 
evident  (Art.  58),  that,  if  these  several 
factors  be  multiplied  together,  they  will 
produce  the  same  product  as  the  origi- 
nal factors,  700  and  40.  Thus  7X4  =  28,  and  28  X  100  =  2800, 
and  2800  X  10  =  28000,  the  same  result  as  in  the  operation. 

Hence,  when  there  are  ciphers,  one  or  more,  on  the  right  of  the 
multiplier,  or  multiplicand,  or  both,  we  may,  for  the  required  pro- 
duct, 

Multiply  the  significant  figures  together,  and  to  their  product  annex 
as  many  ciphers  as  there  are  on  the  right  in  both  multiplicand  and  mul- 
tiplier. 


Examples. 


3.  Multiply  1819  by  10. 

4.  Multiply  4106  by  100. 

5.  Multiply  10000  by  7000. 

6.  Multiply  123000  by  78000. 

7.  Multiply  70000  by  10000. 

8.  Multiply  900900  by  70070. 


Ans.  70000000. 

Ans.  9594000000. 

Ans.  700000000. 

Ans.  63126063000. 


9.  What  must  be  the  distance  sailed  by  a  steamship,  whose 
average  rate  is  310  miles  a  day,  in  making  a  voyage  from  New 
York  to  Liverpool,  in  12  days? 

10.  The  annual  salary  of  a  member  of  Congress  being  3,000 
dollars,  how  much  do  296  members  receive  ? 

Ans.  888,000  dollars. 


44  DIVISION. 

11.  The  salary  of  the  President  of  the  United  States  is 
25,000  dollars  a  year;  how  much  will  it  amount  to  in  82 
years?  Ans.  2,050,000  dollars. 

12.  The  earth  is  95,000,000  of  miles  from  the  sun,  and  the 
planet  Neptune  is  30  times  as  far.  How  far  is  Neptune  from 
the  Sun  ?  Ans.  2,850,000,000  miles. 


DIVISION 


66.  Division  is  the  process  of  finding  how  many  times  one 
number  is  contained  in  another ;  or  the  process  of  separating  a 
number  into  a  proposed  number  of  equal  parts. 

In  division  there  are  three  principal  terms :  the  Dividend, 
the  Divisor,  and  the  Quotient. 

The  dividend  is  the  number  to  be  divided. 

The  divisor  is  the  number  by  which  we  divide 

The  quotient  is  the  result,  or  number  produced  by  the  di- 
vision, and  denotes  the  number  of  times  the  divisor  is  contained 
in  the  dividend,  or  one  of  the  equal  parts  into  which  the  divi- 
dend is  divided. 

When  the  dividend  does  not  contain  the  divisor  an  exact 
number  of  times,  the  excess  is  called  a  remainder,  and  may  be 
regarded  as  a  fourth  term  in  the  division. 

When  the  dividend  consists  of  a  single  denomination,  the 
process  is  termed  Division  of  Simple  Numbers. 

67.  Division  is  frequently  indicated  by  writing  the  dividend 
above  a  short  horizontal  line  and  the  divisor  below ;  thus,  §; 
The  expression  J  =  3  is  read,  6  divided  by  2  is  equal  to  3. 

Another  method  of  indicating  division,  is  by  a  curved  line 
placed  between  the  divisor  and  dividend.  Thus,  the  expression 
6)  12  shows  that  12  is  to  be  divided  by  G. 

68.  When  a  number  is  divided  into  two  equal  parts,  one  of 
the  parts  is  called  one  half ;  when  divided  into  three  equal 
parts,  one  of  the  parts  is  called  one  third,  two  of  the  parts 
two  thirds ;  when  divided  into  four  equal  parts,  one  of  the 


DIVISION.  45 

parts  is  called  one  fourth,  two  of  the  parts  two  fourths,  three 
of  the  parts  three  fourths  ;  etc. 

Such  equal  parts  are  called  fractions,  since  they  are  frac- 
tured or  broken  numbers.  They  are  expressed  by  figures,  in  a 
form  of  division ;  thus,  one  half  is  written  £ ;  one  third,  \  ;  two 
thirds,  § ;  one  fourth,  £ ;  two  fourths,  f ;  three  fourths,  £ ;  and 
may  also  be  read,  one  divided  by  two,  one  divided  by  three, 
and  so  on.  In  any  fraction,  expressed  in  the  manner  now  ex- 
plained, the  number  above  the  line  is  called  the  numerator,  and 
that  below  the  line  its  denominator.  Thus,  in  \,  1  is  the  nu- 
merator, and  2  the  denominator. 

69.  When  the  divisor  and  dividend  are  of  the  same  kind  or 
denomination,  the  quotient  will  denote  the  number  of  times  the 
divisor  is  contained  in  the  dividend.  Thus,  to  find  how  many 
pencils  at  6  cents  each  can  be  bought  for  24  cents,  we  inquire 
how  many  times  6  cents  are  contained  in  24  cents,  which  are 
4  times.  Hence,  4  pencils,  at  6  cents  each,  can  be  bought  for 
24  cents. 

70 •  When  the  divisor  and  dividend  are  of  different  kinds  or 
denominations,  the  divisor  will  denote  the  number  of  equal  parts 
into  which  the  dividend  is  divided,  and  the  quotient  will  denote 
the  number  of  units  in  each  part,  and  will  be  of  the  same  kind 
or  denomination  as  the  dividend.  Thus,  to  find  the  cost  of  1 
pencil,  when  4  pencils  cost  24  cents,  we  separate  or  divide  the 
24  cents  into  4  equal  parts,  of  which  one  part  is  4  cents. 
Hence,  1  pencil  costs  6  cents,  when  4  pencils  cost  24  cents. 

71  #  The  remainder  will  always  be  of  the  same  kind  or  de- 
nomination as  the  dividend,  since  it  is  a  part  of  the  dividend. 

72.  Division  is  the  reverse  of  multiplication.  The  dividend 
answers  to  the  product,  and  the  divisor  and  quotient  to  the 
factors,  of  multiplication.  In  multiplication,  two  factors  are 
given,  to  find  their  product ;  and  in  division,  one  of  two  factors 
and  their  product  are  given,  to  find  the  other  factor. 

73.  To  divide  simple  numbers. 

Ex.  1.  How  many  yards  of  cloth,  at  4  dollars  a  yard,  can 
be  bought  for  948  dollars  ?  Ans.  237  yards. 


46  DIVISION. 

operation.  ^\ye   first  inquire   how  many 

Divisor  4)948  Dividend,  times  4,  the  divisor,  is  contained 

0  o  7   r\     -•     a.  in  9,  the  first  left-hand  figure  of 

1  6  i    quotient. .  the  dlvidendj  wbich  is  ]mn(ireds, 

and  find  it  contained  2  times, 
and  1  hundred  remaining.  We  write  the  2  directly  under  9,  its 
dividend,  for  the  hundreds'  figure  of  the  quotient.  To  4,  the  next 
figure  of  the  dividend,  which  is  tens,  we  regard  as  prefixed  the  1 
hundred  that  was  remaining,  which  equals  10  tens,  and  thus  form 
14  tens,  in  which  we  find  the  divisor  4  to  be  contained  3  times, 
and  2  tens  remaining.  We  write  the  3  for  the  tens'  figure  in  the 
quotient,  and  the  2  tens  that  were  remaining,  equal  20  units,  we 
regard  as  prefixed  to  8,  the  last  figure  of  the  dividend,  which  is  units, 
in  which  the  divisor  4  is  contained  7  times.  Writing  the  7  for  the 
units'  figure  of  the  quotient,  we  have  237  as  the  entire  quotient, 
equal  the  number  of  yards  of  cloth  at  4  dollars  a  yard  that  can  be 
bought  for  948  dollars. 

2.  How  many  times  does  3979  contain  17  ? 

Ans.  234^  times. 

opEnxTioN.  We  say,  17  in  39, 

Dividend.  2  times.     The  2  we 

Divisor  17)3979(234  TU  Quotient.  ™rite  in  the  (luo" 
J  34         V  TT  ^  tient.     17X2  = 

34,  which  we  write 

5  7  under  the  39.     39 

5  1  —  34  =  5,  to  which 

bringing  down  the 

6  9  next  figure  of  the 

6  8  dividend,  we  form 

— ~  -o         .    ,  57.      17   in   57,   3 

1  Remainder.  ^  ^  The  3  \ye 

write  in  the  quo- 
tient. 17  X  3  =  51,  which  we  write  under  the  57.  57  —  51  =  6, 
to  which  bringing  down  the  next  figure  of  the  dividend,  we  form  69. 
17  in  69,  4  times.  The  4  we  write  in  the  quotient.  17  X  4  =  68, 
which  we  write  under  the  69.  69  —  68  =  1,  a  remainder,  or  a  part 
of  the  dividend  left  undivided.  1  divided  by  17  — *  A  (Art.  68). 
The  if  we  write  in  the  quotient,  and  obtain  as  the  answer  required 
234^. 

In  this  illustration,  to  render  the  explanation  the  more  concise,  the 
naming  of  the  denominations  of  the  figures  has  been  omitted. 

When,  as  in  the  operation  preceding  the  last,  results  only  are 
written  down,  the  method  is  called  short  division ;  and  when,  as  in 
the  last  operation,  the  work  is  written  out  at  length,  it  is  called  long 
division.    The  principle  is  the  same  in  both  cases.    Hence  the  general 

Rule.  —  Beginning  at  the  left,  find  how  many  times  the  divisor  is 
contained  in  the  fewest  figures  of  the  dividend  that  will  contain  it,  for 
the  first  quotient  figure. 


DIVISION.  47 

Multiply  the  divisor  by  this  quotient  figure,  and  subtract  the  product 
from  the  figures  of  the  dividend  used.  With  the  remainder,  if  any, 
unite  the  next  figure  of  the  dividend. 

Find  how  many  times  the  divisor  is  contained  in  the  number  thus 
formed,  and  write  the  figure  denoting  the  result  at  the  right  of  the 
former  quotient  figure. 

Thus  proceed  until  all  the  figures  of  the  dividend  are  divided. 

Note  1.  —  The  proper  remainder  is  in  all  cases  less  than  the  divisor.  If,  in 
the  course  of  the  operation,  it  is  at  any  time  found  to  be  larger  than  the  divisor, 
it  will  show  that  there  is  an  error  in  the  work,  and  that  the  quotient  figure 
should  be  increased. 

Note  2. — If  at  any  time  the  divisor,  multiplied  by  the  quotient  figure, 
produces  a  product  larger  than  the  part  of  the  dividend  used,  it  shows  that 
the  quotient  figure  is  too  large,  and  must  be  diminished. 

Note  3.  —  It  will  often  happen  that,  when  a  figure  of  the  dividend  is  taken, 
the  number  will  not  contain  the  divisor ;  and,  in  that  case,  a  cipher  must  be 
placed  in  the  quotient,  and  another  figure  of  the  dividend  taken,  and  so  on, 
until  the  number  is  large  enough  to  contain  the  divisor. 

Note  4.  —  If  there  be  a  remainder  after  dividing  the  last  figure  of  the 
dividend,  write  it  with  the  divisor  underneath,  with  a  line  between  them,  at 
the  right  of  the  quotient. 

74.  First  Method  of  Proof.  —  Multiply  the  divisor  by  the 
quotient,  and  to  the  product  add  the  remainder,  if  any,  and  if 
the  work  be  right,  the  sum  thus  obtained  will  be  equal  to  the 
dividend. 

Note.  —  This  method  follows  from  division  being  the  reverse  of  multiplica- 
tion.   (Art.  72.) 

75.  Second  Method  of  Proof.  —  Find  the  excess  of  nines 
in  the  divisor,  quotient,  and  remainder.  Multiply  the  excess 
of  nines  in  the  divisor  and  quotient  together,  and  to  the  product 
add  the  excess  of  nines  in  the  remainder.  If  the  excess  of  nines 
in  this  sum  equal  the  excess  of  nines  in  the  dividend,  the  work 
may  be  supposed  to  be  right. 

76.  Third  Method  of  Proof.  —  Add  together  the  remainder, 
if  any,  and  all  the  products  that  have  been  produced  by  multi- 
plying the  divisor  by  the  several  quotient  figures,  and  the  result 
will  be  like  the  dividend,  if  the  work  be  right. 

77.  Fourth  Method  of  Proofs — Subtract  the  remainder,  if 
any,  from  the  dividend,  and  divide  the  difference  by  the  quotient. 
The  result  will  be  like  the  original  divisor,  if  the  work  be  right. 

Note.  —  The  first  method  of  proof  (Art.  74)  is  usually  most  convenient, 
and  is  most  commonly  employed. 


48 


DIVISION. 


Examples. 


3.  Divide  9184  by  7. 


Ans.  1312. 


OPERATION. 


Divisor  7)9184  Dividend. 
1312  Quotient. 


4.  Divide  18988  by  759. 


OPERATION. 


PROOF  BY  MULTIPLICATION. 

1312  Quotient. 

7  Divisor. 
9184  Dividend. 

Ans.  25^. 

PROOF    BY    THB    NINES. 

Dividend.  Divisor       =  3  excess. 

Divisor  759)18988(25  Quotient.  Quotient      ==  7  excess. 

1518  Remainder  =  4  excess. 

3808  

3795  (3  X  7)  +  4  =  7  excess. 

1  3  Remainder.  Dividend  =  7  excess. 

5.  Divide  147856  by  97.  Ans.  1524|f 

OPERATION. 

Dividend. 
Divisor  97)147856(1524  Quotient. 
+9  7 

508 

+485 


235 

+19  4 


PROOF  BY  ADDITION. 

97 


416 

+3  8  8 


Remainder. 
14785  6      Dividend. 

Ans.  297. 

PROOF  BY  DIVISION. 

Dividend.  Dividend. 

Divisor  285)84645(297  Quotient.  297)84645(285  Divisor 

570  594 


+2  8  Remainder. 
6.  Divide  84645  by  285. 


OPERATION. 


2764 

2565 

1995 
1995 


2524 
2376 

1485 
1485 


DIVISION. 


49 


7. 
3) 67856336 

22618778 § 

10. 
5)334652 


8. 
7)  178985 


11. 

8)96723578 


9. 
11)1667789 


12. 

9)186731 


13.  14. 

17) 678916(     35)9106013( 

x6.  Divide  671678953  by  6. 

17.  Divide  166336711  by  7. 

18.  Divide  161331793  by  8. 

19.  Divide  161677678  by  9. 

20.  Divide  363895678  by  11. 

21.  Divide  164378956  by  12. 

22.  Divide  78950077  by  3. 

23.  Divide  678956671  by  4. 

24.  Divide  667788976  by  5. 

25.  Divide  777777777  by  6. 

26.  Divide  888888888  by  7. 

27.  Divide  789636  by  46. 

28.  Divide  7967848  by  52. 

29.  Divide  16785675  by  61. 

30.  Divide  675753  by  39. 

31.  Divide  5678911  by  82. 

32.  Divide  6716394  by  94. 

33.  Divide  1167861  by  135. 

34.  Divide  7861783  by  87. 

35.  Divide  1678567  by  365. 

36.  Divide  87635163  by  387. 

37.  Divide  34567890  by  6789. 


15. 
91) 6210011  ( 


Quotients. 

111946492 
23762387 
20166474 
17964186 
33081425 
13698246 


153227 
275175 


71451 
8650 

90365 

4598 

226447 

5091 


38.  What  is  the  value  of  213255467083  -f-  30204  ? 

39.  What  is  the  value  of  395020613  -=-  4444? 

40.  What  is  the  value  of  7207276639  -f-  9009  ? 

41.  What  is  the  value  of  454115186870257  —  500123 

42.  How  many  barrels  of  flour,  at  9  dollars  a  barrel, 
bought  for  18621  dollars? 

N  5 


Rem. 
1. 

2. 
1. 
4. 
3. 
4. 
1. 
3. 
1. 
3. 
6. 

44. 


1. 

Ill, 

28. 

297. 

174. 

5091. 

4267. 

2341. 

4567. 

?  8765. 

can  be 


50  DIVISION. 

43.  How  much  sugar  at  15  dollars  a  hundred  may  be  bought 
for  405  dollars? 

44.  A  tailor  has  938  yards  of  broadcloth ;  how  many  cloaks 
can  be  made  of  the  cloth,  if  it  require  7  yards  to  make  one 
cloak  ? 

45.  What  number  multiplied  by  1728  will  produce  1705536  ? 

Ans.  987. 

46.  A.  Hartmann  has  sold  his  wagon  to  J.  Herr  for  85  dol- 
lars. He  is  to  receive  his  pay  in  wood  at  5  dollars  a  cord. 
How  many  cords  will  it  require  to  pay  for  the  wagon  ? 

Ans.  17  cords. 

47.  The  Bible  contains  31,173  verses;  how  many  must  be 
read  each  day,  that  the  *book  may  be  read  through  in  a  year 
of  365  days?  Ans.  85^|§  verses. 

48.  A  train  on  the  Liverpool  Railroad  runs  at  the  rate  of  65 
miles  an  hour  ;  how  long  would  it  take  at  that  velocity  to  pass 
round  the  earth,  the  distance  being  about  25,000  miles  ? 

Ans.  384T87j  hours. 

49.  A  gentleman  possessing  an  estate  of  66,144  dollars,  be- 
queathed one  fourth  to  his  wife,  and  the  remainder  was  divided 
among  his  4  children ;  what  was  the  share  of  each  ? 

Ans.  12,402  dollars. 

50.  If  the  dividend  is  6756785  and  the  quotient  193051, 
what  is  the  divisor?  Ans.  35. 

51.  A's  age  multiplied  by  17,  or  B's  age  multiplied  by  19, 
is  equal  to  1292  years,  and  the  sum  of  their  ages  is  equal  to 
C's  age  multiplied  by  3.     What  is  the  age  of  each  ? 

Ans.  A's  76  years ;  B's  68  years ;  C's  48  years. 

78.     When  the  divisor  is  a  composite  number. 

Ex.  1.  A  farmer  boughfc  21  horses  for  2625  dollars;  how 
many  dollars  did  each  cost?  Ans.  125  dollars. 

operation.  The  factors  of  21  are 

3)2625  dolls.,  cost  of  21  horses.      3  and   7.     Now,  if  we 

rr  \  q  n  r    -,  11  .     r  rr  -i  divide  the  2625  dollars, 

7  )  875  dolls.,  cost  of  7  horses.       the  cost  of  21  horgeSj  h> 

12  5  dolls.,  cost  of  1  horse.  3,  we  obtain  875  dollars, 

the  cost  of  7  horses,  since 
there  are  7  times  3  in  21.  Then,  dividing  the  875  dollars,  the  cost 
of  7  horses,  by  7,  we  obtain  125  dollars,  the  cost  of  1  horse. 


DIVISION.  51 

2.  Divide  3515  by  42.  Ans.  83£f. 

OPERATION.  FINDING  THE  TRUE  REMAINDER. 

2)3515  4  X  3  X  2  =  24,  1st  Product. 

3  )  17  57,  1,  1st  Rem.  2x2=    4,  2d  Product. 

7)585,  2,  2d  Rem.  1,  1st  Remainder. 

8  3,  4,  3d  Rem.  29,  true  Remainder. 

Or,  (4  X  3  X  2)  +  (2  X  2)  +  1  =  29,  true  Rem. 

Using  as  divisors  2,  3,  and  7,  the  factors  of  42,  we  obtain  for 
remainders,  1,  2,  and  4. 

The  first  remainder,  1 ,  is  a  unit  of  the  given  dividend,  since  it  is  a 
part  of  it  (Art.  71).  The  second  remainder,  2,  is  units  of  the  quo- 
tient 1 75  7,  whose  units  are  2  times  as  great  as  those  of  the  given 
dividend.  The  third  remainder,  4,  is  units  of  the  quotient  585,  whose 
units  are  3  times  as  great  as  those  of  the  quotient  1757,  of  which 
the  units  are  2  times  as  great  as  those  of  the  given  dividend.  Now, 
these  remainders  must  be  all  of  the  same  units  as  the  given  dividend, 
to  constitute  the  whole  or  true  remainder.  We  therefore  multiply 
the  third  remainder  by  3  and  2,  the  divisors  used  in  producing  the 
quotient  of  which  it  is  a  part ;  and  the  second  remainder  by  2,  the 
divisor  used  in  producing  the  quotient  of  which  it  is  a  part ;  and  the 
products  with  the  first  remainder  added  together  give  29,  the  whole 
or  true  remainder  sought.  Hence,  as  shown  by  these  illustrations, 
when  the  divisor  is  a  composite  number,  we  may 

Divide  the  dividend  by  one  of  the  factors,  and  the  quotient  thus  found 
by  another,  and  thus  proceed  till  each  of  the  factors  has  been  made  a 
divisor.     The  last  quotient  will  be  the  quotient  required. 

If  there  be  remainders,  multiply  each  remainder,  except  the  first,  by  all 
the  divisors  preceding  the  one  which  produced  it;  and  the  first  remain- 
der being  added  to  the  sum  of  the  products,  the  amount  will  be  the  true 
remainder. 

Note.  —  There  will  be  but  one  product  to  add  to  the  first  remainder  when 
there  are  only  two  divisors  and  two  remainders. 

Examples. 

3.  Divide  7704  by  24  =  4  X  6.  Ans.  321. 

4.  Divide  8317  by  27  =  3  X  9. 

5.  Divide  3116  by  81  =  9  X  9. 

6.  Divide  61387  by  121  =  11  X  l£  Ans.  507T4fT. 

7.  Divide  19917  by  144  =  12  X  12.  Ans.  138^. 

8.  Divide  91746  by  336  =  6  X  7  X  8.        Ans.  273^- 

9.  At  45  dollars  an  acre,  a  farm  of  how  many  acres  can  be 
bought  for  5464  dollars  ?  Ans.  121£f  acres. 


52  DIVISION. 

79.  When  the  divisor  contains  one  or  more  ciphers  at  the 
right  hand. 

Ex.  1.  If  10  men  receive  792  dollars  for  a  job  of  work, 
what  will  be  each  man's  share  of  it  ?  Ans.  79-£$  dollars. 

operation.  To  multiply  by  10,  we  annex  one 

1  |  0  )  7  9  |  2  cipher,  which  removes  the  figures  one 

r\     &     L  n  c\    o  -r>  place  to  the  left,  and  thus  increases 

Quotient  7  9,  2  Rem.        iheirvaiue  io  times  (Art.  65).   Now, 

Or  thus :  7  9  |  2.  it  is  obvious,  that,  if  we  reverse  the 

process,  and  cut  off  the  right-hand 
figure  by  a  line,  we  remove  the  remaining  figures  one  place  to  the 
right,  and  consequently  diminish  the  value  of  each  10  times,  and  thus, 
divide  the  whole  number  by  10.  The  figures  on  the  left  of  the  line 
are  the  quotient,  and  the  one  on  the  right  is  the  remainder,  which 
may  be  written  over  the  divisor  and  annexed  to  the  quotient  Hence 
each  man's  share  is  79f0. 

2.  How  many  years  will  it  take  a  man  at  a  yearly  salary  of 

700  dollars  to  earn  3664  dollars?  Ans.  5f§-$-  years. 

operation.  The   divisor,  700,  may 

1|00)36|64  be  resolved  into  the  factors 

'  _  -     .   .   ,      _,  7  and  100.   We  first  divide 

7  )  3  6,  6  4, 1st  Kern.  by  thc  factor  100?  by  cut_ 

5,      1,  2d  Rem.      **$  ofr  two  figures  at  the 

right,  and  get  36  for  the 

Or  thus  :   7|00)36|64  quotient,  and  64  for  a  rc- 

mainder.     We  then  divide 

5,  1  G  4.  the   quotient,    36,   by   the 

other  factor,  7,  and  obtain 
5  for  a  quotient  and  1  for  a  remainder.  The  last  remainder,  1,  being 
multiplied  by  the  divisor,  100,  and  64,  the  first  remainder,  added,  we 
obtain  164  for  the  true  remainder  (Art.  77);  and  for  the  answer 
required,  5}oo  years.  Hence,  when  the  divisor  contains  one  or  more 
ciphers  at  the  right,  we  may,  to  perform  the  division, 

Cut  off  the  ciphers  from  the  right  of  the  divisor,  and  the  same  number 
of  figures  from  the  right  of  the  dividend ;  and  then  divide  the  remaining 
figures  of  the  dividend  by  the  remaining  figures  of  the  divisor. 

Note.  —  When,  by  the  operation,  there  is  a  remainder,  to  it  must  be  an- 
nexed the  figures  cut  off  from  the  dividend  to  form  the  true  remainder.  Should 
there  be  no  last  remainder,  then  the  significant  figures,  if  any,  cut  off  from  the 
dividend,  will  form  the  true  remainder. 

Examples. 

Quotient.  Kern. 

3.  Divide  123456789  by  10.        12345678      9. 

4.  Divide  987654300  by  100. 


Quotients. 

Rem. 

5 

2100. 

11 

91853. 

3 

137851. 

37 

411111. 

GENERAL   PRINCIPLES   AND   APPLICATIONS.  53 

5.  Divide  32100  by  6000. 

6.  Divide  3678953  by  326100. 

7.  Divide  1637851  by  500000. 

8.  Divide  41111111  by  1100000. 

9.  Divide  89765432156  by  1000000. 

10.  The  entire  annual  loss  to  the  United  States  in  conse- 
quence of  intemperance  has  been  estimated  to  be  about 
98,400,000  dollars.  How  many  schools  at  a  yearly  expense 
of  600  dollars  would  that  sum  support  ? 

Ans.  164,000  schools. 

11.  The  late  war  with  Russia  was  carried  on  at  a  cost  of 
600,000,000  dollars  to  Great  Britain.  Allowing  that  country 
to  have  a  population  of  28,000,000,  what  was  the  cost  to  each 
individual  ? 

12.  If  light  moves  at  the  rate  of  192,000  miles  in  a  second, 
how  long  is  it  in  passing  from  the  sun  to  the  earth,  a  distance 
of  95,000,000  of  miles.  Ans.  494|f  §  seconds. 


GENERAL  PRINCIPLES  AND  APPLICATIONS. 

80 1  In  division,  the  value  of  the  quotient  depends  upon  the 
divisor  and  dividend. 

81.  If  the  dividend  be  multiplied,  or  the  divisor  divided,  by 
any  number,  the  quotient  is  multiplied  by  the  same  number. 
Thus,  if  the  dividend  be  20  and  the  divisor  4,  the  quotient 
will  be  5 ;  but  if  the  dividend  be  multiplied  by  any  number, 
as  2,  and  the  divisor  remain  unchanged,  the  quotient  will  be  2 
times  as  large  as  before,  or  10;  as  (20  X  2)  ~  4  =  10;  and 
if  the  divisor  be  divided  by  the  2,  and  the  dividend  remain  un- 
changed, the  quotient  will  be,  likewise,  2  times  as  large,  or  10 ; 
as  20 -*•  (4  -f-  2)=~  10. 

82.  If  the  dividend  be  divided,  or  the  divisor  multiplied,  by 
any  member,  the  quotient  is  divided  by  the  same  number.  Thus, 
if  the  dividend  be  32  and  the  divisor  8,  the  quotient  will  be  4. 
But  if  the  dividend  be  divided  by  any  number,  as  2,  and  the 

5* 


54  CANCELLATION. 

divisor  remain  unchanged,  the  quotient  will  be  only  half  as 
large  as  before,  or  2  ;  as  (32  -H  2)  -r-  8=2;  and  if  the 
divisor  be  multiplied  by  2,  and  the  dividend  remain  unchanged, 
the  quotient  will  be,  likewise,  only  half  as  large,  or  2  ;  as  32  -f- 
(8  X  2)  =  2. 

83.  If  the  dividend  and  divisor  be  both  multiplied,  or  both 
divided,  by  the  same  number,  the  quotient  will  not  be  changed. 
Thus,  if  the  dividend  be  16  and  the  divisor  4,  the  quotient  will 
be  4.  Now,  if  we  multiply  the  dividend  and  divisor  by  some 
number,  as  2,  their  relative  values  are  not  changed,  and  we 
obtain  32  and  8  respectively,  and  32  -r-  8  =  4,  the  same  as 
the  original  quotient.  Also,  if  we  divide  the  dividend  and 
quotient  by  some  number,  as  2,  their  relative  values  are  not 
changed,  and  we  obtain  16  and  2  respectively,  and  16  -s-  2  =  8, 
the  same  quotient  as  before. 

84.  If  a  factor  in  any  number  is  rejected  or  cancelled,  the 
number  is  divided  by  that  factor.  Thus,  if  24  is  the  dividend 
and  6  the  divisor,  the  quotient  will  be  4.  Now,  since  the  di- 
visor and  quotient  are  the  two  factors  which,  being  multiplied 
together,  produce  the  dividend  (Art.  72),  it  follows,  if  we  reject 
or  cancel  the  factor  6,  the  remaining  4  is  the  quotient ;  and,  by 
the  operation,  the  dividend  24  has  been  divided  by  4. 

CANCELLATION. 

85.  Cancellation  is  the  method  of  abbreviating  arithmet- 
ical operations  by  rejecting  any  factor  or  factors  common  to  the 
divisor  and  dividend. 

Ex.  1.  Sold  19  thousand  shingles  at  4  dollars  a  thousand, 
and  received  pay  in  wood  at  4  dollars  a  cord ;  how  many  cords 
of  wood  was  received?  Ans.  19  cords. 

operation.  Having  indicated  by  signs 

Dividend    ^  X  19  .  the  multiplication  and  di- 

.  ~Z        =s  1™  Quotient,      vision  required  by  the  ques- 

Divisor  £  tjon^   ^en^   since    dividing 

both  dividend  and  divisor 
by  the  same  number  will  not  change  the  quotient  (Art.  83),  we  divide 
them  by  the  common  factor  4,  by  cancelling  it  in  both,  and  obtain  25 
for  the  quotient. 


CANCELLATION.  55 

2.  Divide  the  product  of  15,  3,  28,  and  13,  by  the  product 

of  7,  30,  and  4.  * 

OPERATION. 

Dividend  j  $  X  3  X  ft  0  X  1  3  _    3  9 

Divisor  *X£0Xrf~  "  *  T  =  19*  Qu0tient' 

2 

The  product  of  the  7  and  4  in  the  divisor  equals  the  28  in  the  divi- 
dend ;  we  therefore  cancel  all  these  numbers.  Finding  15  in  the 
dividend  to  be  a  factor  of  30  in  the  divisor,  we  cancel  both  of  the 
numbers,  and  use  the  remaining  factor  2  in  place  of  the  30.  There 
now  being  no  factor  common  to  both  dividend  and  divisor  uncancelled, 
we  multiply  together  the  remaining  factors  in  the  dividend,  and  divide 
the  product  by  the  remaining  factor  in  the  divisor,  and  obtain  the 
quotient  19J. 

Rule.  —  Cancel  the  factor  or  factors  common  to  the  dividend  and 
divisor,  and  then  divide  the  product  of  the  factors  remaining  in  the 
dividend  by  the  product  of  those  remaining  in  the  divisor. 

Note.  —  1.  In  arranging  the  numbers  for  cancellation,  the  dividend  may  be 
written  above  the  divisor,  with  a  horizontal  line  between  them,  as  in  division 
(Art.  67);  or,  as  some  prefer,  the  dividend  may  be  written  on  the  right  of  the 
divisor,  with  a  vertical  line  between  them. 

Note.  —  2.  Cancelling  a  factor  does  not  leave  0,  but  the  quotient  1,  to  take 
its  place,  since  rejecting  a  factor  is  the  same  as  dividing  by  that  factor  (Art. 
84).  Therefore,  for  every  factor  cancelled,  either  in  the  dividend  or  divisor, 
the  factor  1  remains. 

Examples. 

3.  Multiply  24  by  16,  and  divide  the  product  by  12. 

Ans.  32. 

4.  Divide  48  by  16,  and  multiply  the  quotient  by  8. 

Ans.  24. 

5.  Divide  the  product  of  7,  10,  12,  and  5,  by  the  product  of 
14,  18,  and  6. 

6.  If  15  be  multiplied  by  7,  27,  and  40,  and  the  product 
divided  by  54  multiplied  by  14,  10,  and  2,  what  will  be  the 
result  ?  Ans.  7£. 

7.  Divide  the  product  of  13, 15,  20,  and  5,  by  the  product  of 
26,  10,  2,  and  3.  Ans.  121. 

8.  Divide  the  product  of  28,  27,  21,  15,  and  18,  by  the  pro- 
duct of  7,  54,  7,  3,  and  9. 

9.  How  many  pounds  of  butter  at  28  cents  a  pound  will 
be  required  to  pay  for  56  pounds  of  sugar  at  11  cents  a 
pound  ?  Ans.  22  pounds. 


56  CONTRACTIONS   IN   MULTIPLICATION. 

10.  A.  Holmes  sold  14  boxes  of  soap,  each  containing  24 
pounds,  at  9  cents  a  pound,  and  received  for  pay  63  barrels  of 
ashes,  each  containing  3  bushels.  What  was  allowed  a  bushel 
for  the  ashes  ?  Ans.  1 6  cents. 

11.  M.  Gardner  sold  5  piles  of  brick,  each  containing  12 
thousand,  at  7  dollars  a  thousand,  and  was  paid  in  wood,  3 
ranges,  at  4  dollars  a  cord.     How  many  cords  in  each  range  ? 

Ans.  35  cords. 

12.  A  merchant  exchanged  8  cases  of  shoes,  each  containing 
60  pairs,  at  75  cents  a  pair,  for  a  certain  number  of  casks  of 
molasses,  each  containing  90  gallons,  at  40  cents  a  gallon. 
How  many  casks  did  he  get  ?  Ans.  10  casks. 

CONTRACTIONS    IN    MULTIPLICATION. 

86 •  A  contraction  is  the  process  of  shortening  any  oper- 
ation. 

87.  When  the  multiplier  is  13,  14,  etc.,  or  1  with  a  sig- 
nificant figure  annexed. 

Ex.  1.  Multiply  3126  by  14.  Ans.  43764. 

first  operation.         second  operation.  In  the  first  operation, 

3126X14  3126  we  multiply  the  multi- 

12  5  0  4  14  plicand   by   the   4  units 

.  0  _  „  .     .  ,  0  _      •,  of    the    multiplier,   and 

4  3  7  6  4  Ans.  4  3  7  6  4  Ans.     write  tlie   product   un_ 

der  the  multiplicand  one 
place  to  the  right.  To  this  partial  product  we  add  the  multiplicand, 
since,  as  it  stands,  it  represents  the  product  of  the  multiplicand  by 
the  1  ten  of  the  multiplier ;  and  obtain  43764,  the  answer  required.  In 
the  second  operation,  we  add  in  the  multiplicand  taken  as  the  product 
by  the  1  ten  of  the  multiplier,  as  we  multiply  by  the  4  units ;  thus, 
6  X  4  =  24,  of  which  we  write  down  the  4  and  carry  the  2. 
2X4  =  8,4-2  (carried)  +  6  (from  the  multiplicand)  =  16,  of 
which  we  write  down  the  6  and  carry  the  1.  lX4  =  4,-j-l-|- 
2=7,  which  we  write  down.  3X4  =  1 2, -J-l  =13,  of  which  we 
write  down  the  3  and  carry  the  1.  3  — [—  1  =  4,  which  we  write 
down  ;  and  have  as  the  entire  result  43764  as  before. 

Rule.  —  Write  the  product  by  the  units'  figure  of  the  mvltiplier 
under  the  midtijrficand,  one  place  to  the  right,  and  add  them  together. 
Or, 

Multiply  each  figure  of  the  multiplicand  "by  the  units1  figure  of  the 
multiplier,  and,  after  the  units'  place,  add  in  the  preceding  figure  of 
the  midtiplicand. 


CONTRACTIONS   IN   MULTIPLICATION.  57 

Examples. 

2.  Multiply  63013  by  17.  Ans.  1071221. 

3.  Multiply  79245  by  19. 

4.  Multiply  32067812  by  16.  Ans.  513084992. 

88.  When  the  multiplier  is  101,  102,  etc.,  or  1  with  one  or 
more  ciphers  and  a  significant  figure  annexed. 

Ex.  1.   Multiply  8107  by  103.  Ans.  835021. 

operation.  We  multiply  by  3,  the  units'  figure  of  the 

8107X103       multiplicand,  and  write  the  product  under 

2  4  3  2  1  the  multiplicand,  two  places  to  the  right, 
o  o  g  n  9  i     a  so  *^afc  tne  multiplicand,  as  it  stands  over 
o  o  0  0  Z  1  Ans.      thig  partial  product,  will  represent  the  pro- 
duct of  the  multiplicand  by  the  1  hundred 

of  the  multiplier ;  and,  adding  these,  we  obtain  835021,  the  result  re- 
quired. For  the  reason  given,  if  the  multiplier  had  been  such  as  to 
have  contained  one  more  intervening  cipher,  we  should  have  written 
the  product  by  the  units'  figure  three  places  to  the  right,  and  so  on, 
one  place  farther  to  the  right  for  every  additional  intervening  cipher. 

Rule.  —  Write  the  product  by  the  units7  figure  of  the  multiplier 
under  the  multiplicand,  as  many  places  to  the  right  as  there  are  in  the 
multiplier  intervening  ciphers  plus  1 ;  and  add  them  together. 

Examples. 

2.  Multiply  6651  by  108.  Ans.  718308. 

3.  Multiply  111223  by  104.  Ans.  11567192. 

4.  Multiply  2042  by  1009. 

89  #  When  the  multiplier  is  21,  31,  etc.,  or  1  with  a  signifi- 
cant figure  prefixed. 

Ex.  1.  Multiply  3113  by  41.  Ans.  127633. 

operation.  \ye  multiply  by  4,  the  tens'  figure  of  the 

3  113X^1       multiplier,  and  write  the  product  under  the 
12  4  5  2  multiplicand,  one  place  to  the  left,  so  that 

9  _  the  multiplicand,  as  it  stands  over  this  par- 

1  L  4  b  o  o  Ans.         tial  product,  will  represent  the  product  of 
the  multiplicand  by  the  1  unit  of  the  multi- 
plier ;  and,  adding  these,  obtain  the  result  required. 

Rule.  —  Write  the  product  by  the  tens9  figure  of  the  multiplier 
under  the  multiplicand,  one  place  to  the  left,  and  add  them  together. 


58  CONTRACTIONS   IN   MULTIPLICATION. 

Examples. 

2.  Multiply  13317  by  51.  Ans.  679167. 

3.  Multiply  71389  by  21. 

4.  Multiply  12062  by  91.  Ans.  1097642. 

90.  When  the  multiplier  is  201,  301,  etc.,  or  1  with  one  or 
more  ciphers  and  a  significant  figure  prefixed. 

Ex.  1.  Multiply  14118  by  601.  Ans.  8484918. 

operation.  We  multiply  by   6,   the    hundreds' 

14118  X  601.      figure  of  the  multiplier,  and  write  the 

8  4  7  0  8  product  under    the   multiplicand,  two 

places  to  the  left,  so  that  the  multipli- 
ed 4  o  4  J  1  o  Ans.  cand,  as  it  stands  over  this  partial  pro- 
duct, will  represent  the  product  of  the 
multiplicand  by  the  1  unit  of  the  multiplier ;  and  adding  these  we 
have  the  answer  required. 

Rule.  —  Write  the  product  by  the  hundreds'  figure  of  the  multiplier 
under  the  multiplicand,  as  many  places  to  the  left  as  there  are  in  the 
multiplier  intervening  ciphers  plus  1 ;  and  add  them  together. 

Examples. 

2.  Multiply  8360  by  7001.  Ans.  58528360. 

3.  Multiply  10613  by  801.  Ans.  8501013. 

4.  Multiply  91603  by  2001. 

91.  When  the  multiplier  or  multiplicand  has  a  fraction 
annexed. 

Ex.  1.  Multiply  426  by  7£.     By  8§.     Ans.  3124 ;  3692. 

OPERATION.  OPERATION. 

426  426 

7*  8f 

2  9  8  2  =  Product  by  7.  3408  =  Product  by  8. 

14  2=  Product  by  £.  284  =  Product  by  §. 

3124  =  Product  by  7 J.  3  6  9  2  =  Product  by  8f. 

In  multiplying  426  by  7^,  we  first  obtain  the  product  of  426  by  7, 
and  then  the  product  of  426  by  |,  and,  adding  these  two  partial  pro- 
ducts together,  have  3124,  the  product  by  7£.  In  multiplying  by  £, 
we  take  one  third  of  the  multiplicand,  by  dividing  it  by  3  ;  thus, 
426  x  ^  =  426  -7-  3  =  142.  In  multiplying  426  by  8|,  we  proceed 
as  in  the  other  case,  except  in  obtaining  the  product  by  the  fraction  ; 


CONTRACTIONS   IN   MULTIPLICATION.  59 

we  take  two  thirds  of  the  multiplicand,  by  taking  one  third  of  it  2 
times  ;  thus,  426  X  t  =  (426  -f-  3)  x  2  =  284. 

If  the  fraction  had  been  annexed  to  the  multiplicand  instead  of 
the  multiplier,  we  then  would  have  found  the  product  of  the  frac- 
tion and  multiplier,  for  a  partial  product,  in  like  manner  as  above. 

Rule.  —  Multiply  the  fractional  part  and  the  whole  number  sepa- 
rately, and  add  the  products. 

Examples. 

2.  Multiply  915  by  22§.  Aiis.  20496. 

3.  Multiply  1224|  by  18.  Ans.    22034f 

4.  If  69 £  miles  make  1  degree,  how  many  miles  are  180 
degrees  ?  Ans.  12450  miles. 

92.     When  the  multiplier  is  a  convenient  part  of  a  number 

of  tens,  hundreds,  or  thousands. 

Note.  —  The  following  are  some  of  the  convenient  parts  often  occurring  as 
multipliers;  2£  =%  of  10;  3£  =  £  of  10;  12£  =  £  of  100;  16§  =  i  of  100; 
25  =  £  of  100;  33£  =  £  of  100;  125  =£  of  1000;  166|  =  £of  1000;  250  =  £ 
of  1000;  333^  =  J  of  1000. 

Ex.  1.  Multiply  785643  by  25.  Ans.  19641075. 

operation.  We  multiply  by  1 00,  by  annex- 

4)78564300  ing  two  ciphers  to  the  multipli- 

i  n  n  a  i  r\  n  z   -n     i  cand  (Art.  65),  and  obtain  a  pro- 

19  6  410  7  5  Product.     duct |  times  £  Urge  a?  it  s  Juld 

be,  since  25,  the  multiplier,  is  only 
one  fourth  part  of  100;  we  therefore  take  one  fourth  part  of  that 
product,  by  dividing  by  4,  for  the  true  product. 

Upon  the  same  principle,  if  the  multiplier  had  been  3£,  we  should 
have  annexed  one  cipher  and  divided  by  3,  or,  if  the  multiplier  had 
been  125,  we  should  have  annexed  three  ciphers  and  divided  by  8. 
Hence  the  following 

Rule.  —  Multiply  by  the  number  of  tens,  hundreds,  or  thousands,  of 
which  the  multiplier  is  a  part,  and,  of  the  product  thus  found,  take  the 
same  part. 

Examples. 

2.  Multiply  68056  by  12£.  Ans.  850700. 

3.  Multiply  17924  by  2£. 

4.  Multiply  192378  by  16§.  Ans.  3206300. 

5.  Multiply  12345678  by  125.  Ans.  1543209750. 

6.  How  much  can  be  earned  in  one  year  of  313  working 
days,  at  3£  dollars  a  day  ? 


60  CONTRACTIONS   IN  MULTIPLICATION. 

7.  What  will  a  farm  containing  534  acres  cost,  at  33£  dollars 
an  acre  ? 

8.  In  a  certain  large  field  there  are  771  rows  of  corn,  each 
containing  250  hills  ;  how  many  hills  in  all  ? 

Ans.  192750  hills. 

9.  From  a  port  in  Louisiana  there  were  exported,  in  a  given 
time,  9168  boxes  of  sugar,  averaging  166§  pounds  to  a  box; 
required  the  whole  number  of  pounds. 

Ans.  1528000  pounds. 

10.  An  agent  has  bought  for  the  army,  at  different  times,  in 
the  aggregate,  1993  horses,  at  an  average  price  of  125  dollars 
each.     How  much  did  he  pay  for  them  in  all  ? 

Ans.  249125  dollars. 

11.  How  many  are  333£  times  28044  ?       Ans.  9348000. 

93.  When  a  part  of  the  multiplier  is  a  factor  of  another 
part. 

Ex.  1.    Multiply  3263  by  568.  Ans.  1853384. 

operation.  ^  "We  regard  the  multiplier 

3  2  6  3  Multiplicand.  as  separated  into  two  parts, 

5  6  8  Multiplier.  56  tens  and  8  units,  or  560 

_.  _  .  _   .         .~     ,        .      B       .  -(-  8 ;  of  which  the  smaller 

2  6  10  4=  Product  by  8  units.  part  is  evidently  a  factor  of 

18  2  7  2  8     =  Product  by  56  tens,  the  larger,  since  the  56  tens, 

l85lT^l  =  Product  by  568.         ^^^ft  8 

units,  obtaining  the  product 
for  that  part  of  the  multiplier.  Now,  as  this  product  is  the  same  as 
that  by  the  factor  8  of  the  other  part  of  the  multiplier,  we  multiply  it 
by  7  tens,  obtaining  the  product  of  the  multiplicand  by  8  X  7  tens  or 
56  tens.  These  products  of  the  parts,  560  and  8,  added  together,  give 
the  true  product  by  568. 

Rule.  —  Multiply  first  by  the  smaller  part  of  the  multiplier;  and  then 
that  partial  product  by  a  factor,  or  factors,  of  a  larger  part ;  and  soon 
with  all  the  parts.  The  sum  of  the  several  partial  products  ivill  be  the 
product  required. 

Note.  —  Care  must  be  taken  in  writing  down  the  partial  products  to  have 
the  units  of  the  different  orders  stand  in  their  proper  places  for  adding. 

Examples. 

2.  Multiply  112345678  by  288144486. 

Ans.  32371787641631508. 


CONTRACTIONS   IN    MULTIPLICATION.  61 


OPERATION. 


112345  6  78  Multiplicand. 
288144486  Multiplier. 

674074068  =  Product  by  6  units. 

5392592544        =    1st  product  X  8  tens  for  product  by  48  tens. 
1    (\  1    77777fi^2  2(*  Product  x  3  thousands  for  product  by  144 

32355555264  =3d  *fiS2£*  millions  for  product  by  m 

32~3  71787641631508  =  Product  by  288144486. 

3.  Multiply  61370913  by  96488.     Ans.  5921556653544. 

4.  Multiply  8649347864  by  1325769612. 

Ans.  11467042561708308768. 

94.    When  the  multiplier  is  any  number  of  nines. 

Ex.  1.     Multiply  87654  by  999.  Ans.  87566346. 

t-  operation.  By  annexing  three  ci- 

87654000  =  87654X1000     phers  to  the  multiplicand 
87654  =  87654X  1     we  .take  &  1000  times,  or 

— -  „  ~  ~  ^  .  n        r,  -  n  k  .  ^  ~  ~      1  time  more  than  is  re- 

87566346  =  87654X     999     quired  by  the  given  mul- 

tiplier.  We  therefore  from 
this  result  subtract  the  multiplicand  taken  once,  and  thus  obtain  the 
product  by  999.  In  like  manner  we  may  multiply  by  any  number  of 
nines.     Hence  the 

Rule.  —  Annex  as  many  ciphers  to  the  multiplicand  as  there  are 
nines  in  the  multiplier,  and  from  the  number  thus  produced  subtract 
the  given  multiplicand. 

Note.  —  To  multiply  by  any  number  of  threes,  find  the  product  for  the  same 
number  of  nines,  by  the  rule,  and  take  one  third  of  it  by  dividing  by  3 ;  and  to 
multiply  by  any  number  of  sixes,  take  twice  the  product  of  the  same  number 
of  threes,  by  multiplying  it  by  2. 

2.  Multiply  7777777  by  9999.  Ans.  77769992223. 

3.  Multiply  416231  by  99999. 

4.  Multiply  987654  by  333333.  Ans.  329217670782. 

5.  Multiply  876543  by  mm.  Ans.  58435615638. 

6.  Multiply  999999  by  9999.  Ans.  9998990001. 

7.  Multiply  32567895  by  3333. 

8.  Multiply  66666  by  66666.  Ans.  4444355556. 

9.  Multiply  912345678  by  99.  Ans.  90322222122. 

10.  Multiply  1234567  by  9999.  Ans.  12344435433. 

11.  Multiply  98123452  by  999999.  Ans.  98123353876548. 
N  6 


62  CONTRACTIONS   IN    DIVISION. 


CONTRACTIONS  IN  DIVISION. 

95.  When  the  divisor  is  a  convenient  part  of  a  number  of 
tens,  hundreds,  or  thousands. 

Ex.  1.     Divide  19641075  by  25.  Ans.  785643. 

operation.  By   multiplying   both   divisor  and 

1964107  5  dividend  by  4,  which  does  not  change 

4  the  relation  of  the  one  to  the  other 

WfttAiftlvA    -       .  (Art.   83),  the  divisor  becomes  100, 

785643100  Quotient.  which  enables  us  to  perform  the  di- 
vision by  simply  cutting  off  two  figures 
at  the  right  of  the  dividend  (Art.  79).  In  like  manner,  if  the  divisor 
had  been  3£,  by  taking  both  it  and  the  dividend  3  times  as  large,  we 
could  have  performed  the  division  by  simply  cutting  off  one  figure  at 
the  right  of  the  dividend ;  or,  if  the  divisor  had  been  125,  by  taking 
both  it  and  the  dividend  8  times  as  large,  we  could  have  performed  the 
division  by  simply  cutting  off  three  figures  at  the  right  of  the  dividend. 
Hence  the 

Rule.  —  Multiply  both  divisor  and  dividend  by  that  number  which 
?/•///  change  the  divisor  to  a  number  of  tens,  hundreds,  or  thousands,  and 
then  divide. 

Examples. 

2.  Divide  89630  by  31.  Ans.  26889. 

3.  Divide  123450  by  16§. 

4.  Divide  18621  by  12£.  Ans.  1489-^. 

5.  Divide  317121  by  2J.  Ans.  126848T\. 

6.  Divide  876735  by  33£.  Ans.  26302T^. 

7.  Divide  123456  by  125. 

8.  Divide  61678500  by  250.  Ans.  246714. 

9.  J.  Cushing  bought  a  number  of  horses,  at  166f  dollars 
each,  for  $  9500  ;  required  the  number  bought. 

Ans.  57  horses. 

10.  A  company  has  received  12000  dollars  from  the  sale  of 
piano-fortes,  at  an  average  price  of  333^  dollars  each ;  required 
the  number  sold. 

11.  How  many  shares  of  the  Illinois  Central  Railroad,  at 
125  dollars  each,  can  be  bought  for  150000  dollars  ? 

Ans.  1200  shares. 

12.  How  many  cows,  at  33£  dollars  each,  may  be  purchased 
for  333£  dollars. 


CONTRACTIONS   IN   DIVISION.  63 

13.  How  many  books,  at  2£  dollars  each,  may  be  purchased 
for  120  dollars  ?  Ans.  48  books. 

14.  A  certain  magazine  contains  616350  pounds  of  powder, 
in  kegs  of  25  pounds  each ;  required  the  number  of  kegs. 

Ans.  24654  kegs. 

96.    When  the  divisor  is  any  number  of  nines. 

Ex.  1.     Divide  316234  by  99.  Ans.  3l94f|. 

operation.  vye  first  divide  by  100,  or  99  -f-  1,  by  cut- 

3  16  2  3  4  ting  off  two  figures  at  the  right  of  the  dividend, 

3  19  6  obtaining  for  the  first  partial  quotient  3162 

12  7  and  a  remainder  34.     Since  the  divisor  used 

2  g  was  1  larger  than  the  given  divisor,  99,  the 

quotient  obtained  denotes  that  an  excess  of 

3  1  9  4  f  §-  Ans.  3162,  or  a  number  equal  to  the  quotient  itself, 
must  be  added  to  the  34  for  the  true  remainder, 
34  _|_  3162  =  3196,  which  exceeding  the  divisor  99,  we  write  it  for  a 
second  dividend;  and  dividing  by  100,  or  99  -f-  1,  as  before,  we  ob- 
tain for  the  second  partial  quotient  31,  and  a  remainder  96.  To  the 
remainder  96  we  add  31,  the  excess  denoted  by  the  last  quotient, 
and  obtain  1 2  7  for  a  third  dividend,  which  being  divided  by  1 00,  or 

99  -j-  1  gives  the  third  partial  quotient  1,  and  a  remainder  27.  To 
the  remainder  27  adding  1,  the  excess  denoted  by  the  last  quotient, 
we  have  for  the  true  remainder  28,  which  is  ||.  The  sum  of  the 
partial  quotients  with  the  final  remainder  annexed  gives  3194||,  the 
quotient  required. 

The  above  process  is  based  upon  the  principle  that  10  =  9  -f-  1 ; 

100  =  99  -f-  1 ;  1000  =  999  +  1,  etc.  ;  consequently  20  =  (2  X  9) 
+  2,  and  37  =  (3  X  9)  -f-  3  -f  7 ;  200  ==  (2  X  99)  -f  2 ;  3859  = 
(38  X  99)  +  38  +  59  ;  15987  =  (15  X  999)  -f  15  -f  987,  etc. 
Hence*  316234  =  316200  -j-  34  =  (3162  X  99)  -f-  3162  -f  34; 
3162  -f  34  ==  3196  ==  3100  -f-  96  =  (31  X  99)  +  31  -f-  96  ;  31  -f- 
96  =  127  =  100  +  27=  (1  X  99)  +1  +  27;  1  -f  27  =  28  =  jj; 
and  3162  — [—  31  — |—  1  — }—  11  ==  3194||,  the  answer,  as  before  obtained. 
By  like  process,  and  upon  the  same  principle,  may  the  quotient  be 
found  for  any  number  of  nines. 

Rule.  —  Add  1  to  the  given  divisor,  and,  by  it  thus  increased,  di- 
vide by  Gutting  off  figures  at  the  right  of  the  dividend.  To  the  figures 
cut  off  on  the  right  add  those  on  the  left  for  a  true  remainder,  of  which, 
if  it  equal  or  exceed  the  given  divisor,  make  a  second  dividend,  and  divide 
as  before.  Proceed  thus  till  there  shall  be  no  remainder  as  large  as  the 
given  divisor ;  and  the  sum  of  the  several  quotients,  with  the  last  remain- 
der, if  any,  will  be  the  answer  required. 

Note.  —  When  the  last  remainder  is  the  same  as  the  given  divisor,  it  must 
be  cancelled,  and  1  written  as  a  partial  quotient. 


64  CONTRACTIONS   IN   DIVISION. 

Examples. 
2.  Divide  341  by  9.  3.  Divide  123332544  by  999. 

OPERATION.  OPERATION. 

544 


34 
3 


1  123332 

5  123 

8  1 


876 
000 


3  7  f  Ans.  12  3  4  5  6  Ans. 

4.  Divide  12332655  by  999.  Ans.  12345. 

5.  Divide  987551235  by  9999. 

6.  Divide  9123456779876543211  by  999999999. 

Ans.  9123456789, 

97.     Abbreviated  method  of  long  division. 

Ex.  1.     Divide  34634  by  134.  Ans.  258T%. 

operation.  The  operation  we  perform 

134)34634(258  ^&  Ans.     thus :  finding  the  first  quotient 

„  Q  cy  figure  to  be  2,  we  say  4  X  2  = 

'  °6  8;  16  —  8  =  8;  3X2  =  6, 

113  4  and  1  carried  =  7  ;  14  —  7 

=  7 ;  1X2  =  2,  and  1  car- 

6  2  ried  =  3 ;  3  —  3  =  0.     We 

now  bring  down  3,  and  find 
the  next  quotient  figure  to  be  5,  then,  4X5  =  20;  3  —  0  =  3; 
3  X  5  =  15,  and  2  carried  =  17;  8  —  7=1;  1x5  =  5,  and  1 
carried  =  6 ;  7  —  6=1.  We  next  bring  down  4,  and  find  the  next 
quotient  figure  to  be  8 ;  then,  4X8  =  32  ;4  —  2=2;  3X8  = 
2  I,  and  3  carried  =  27  ;  13  —  7  =  6  ;  1  X  8  =  8,  and  2  -f  1  car- 
ried =  11;  1  —  1  =  0;  1  —  1  =  0.  Hence,  this  method  is  that  of 
ordinary  long  division  (Art.  73),  abridged  by  subtracting  each  figure 
of  the  product  of  the  divisor  and  a  quotient  figure,  as  it  is  obtained,  and 
writing  down  onbj  the  remainders. 

Examples. 

2.  Divide  39006  by  44.  Ans.  886f  |. 

3.  A  certain  orchard  contains  1088  trees  in  34  rows ;  how 
many  trees  in  each  row  ?  Ans.  32  trees. 

4.  A  speculator  sold  191  mules  at  a  gain  of  5157  dollars; 
what  was  the  gain  on  each  ? 

5.  The  population  of  Massachusetts,  in  1855,  was  1,133,123 ; 
how  many  would  that  be  to  each  of  its  7750  square  miles  of 
surface  ?  Ans.  146|f  §  g. 


8# 


CONTRACTIONS   IN   DIVISION.  65 

98.    When  the  divisor  or  dividend  has  a  fraction  annexed. 
Ex.  1.     Divide  3692  by  8|.  Ans.  426. 

OPERATION. 

3  6  9  2  We  bring  the  divisor  and  divi- 
3  dend  to  the  same  fractional  parts 
— —  as  those  denoted  by  the  given 
26)11076(426  Ans.  fraction,  which  are  thirds,  by  mul- 
10  4  tiplying  them  both  by  3,  the  num- 
7TZ  ber  of  thirds  in  1.  We  thus  change 
r  9  the  divisor  to  26  thirds,  and  the 
0  ^  dividend  to  3692  thirds,  without 
15  6  changing  the  quotient  (Art.  83)  ; 
i  k  a  and  dividing  obtain  426,  the  an- 
swer  required. 

Rule.  —  Reduce  the  divisor  and  dividend  to  the  same  fractional 
parts  as  are  denoted  by  the  given  fraction,  and  then  divide  as  in  whole 
numbers. 

Note.  —  In  case  there  should  be  a  remainder  after  the  division,  its  true 
value  may  be  found  as  in  Art.  76. 

Examples. 

2.  Divide  13120  by  9f  Ans.  1418£f 

3.  Divide  76672f  by  36.  Ans.  2129f  ||. 

4.  Divide  2090£  by  205.  Ans.  10£f£. 

5.  If  16J  feet  make  a  rod,  how  many  rods  are  there  in  10626 
feet  ? 

6.  If  69£  degrees  make  a  mile,  how  many  degrees  in  12450 
miles  ?  Ans.  180  degrees. 

7.  How  many  barrels  of  pork,  at  17f  dollars  a  barrel,  may  be 
bought  for  559 1|-  dollars  ? 

8.  A  merchant  has  sold  2667  yards  of  silk  in  dress  patterns 
of  10^-  yards  each ;  required  the  number  of  patterns. 

Ans.  254  patterns. 

9.  How  many  plats,  each  of  272J  square  feet,  are  equal  in 
extent  to  136125  square  feet  ?  Ans.  500  plats. 

10.  How  many  days  will  119  pounds  of  bread  last  a  man,  if 
he  consume  2^  pounds  per  day  ?  Ans.  49  days. 

11.  How  many  casks  of  31£  gallons  each  will  be  required  to 
hold  12968J  gallons  of  cider?  Ans.  415  casks. 

G* 


66  PROBLEMS. 

PKOBLEMS, 

FOUNDED  UPON  THE  FUNDAMENTAL  RULES. 

99.  The  following  problems  are  founded  upon  the  general 
principles  of  addition,  subtraction,  multiplication,  and  division, 
the  fundamental  operations  of  arithmetic,  which  have  already 
been  explained. 

1.  The  parts  of  a  number  being  given,  to  find  the  number.  — 
Add  the  parts  together  (Art.  47). 

2.  The  sum  of  two  numbers  and  one  of  the  numbers  beinjr 
given,  to  find  the  other  number.  —  From  the  sum  subtract  the 
given  number  (Art.  50). 

3.  The  difference  between  two  numbers  and  the  larger  num- 
ber being  given,  to  find  the  smaller.  —  From  the  larger  number 
subtract  the  difference  (Art.  52). 

4.  The  difference  between  two  numbers  and  the  smaller 
number  being  given,  to  find  the  larger.  —  Add  the  smaller 
number  and  the  difference  together  (Art.  51). 

5.  The  sum  and  the  difference  of  two  numbers  being  given, 
to  find  the  numbers.  —  From  the  sum  subtract  the  difference  and 
divide  the  remainder  by  2,  for  the  smaller  number ;  add  the 
difference  to  the  smaller  number,  for  the  larger  (Art.  52). 

Note.  —  In  like  manner,  when  the  sum  and  differences  are  given,  may  be 
found  any  number  of  required  numbers.  After  subtracting  the  differences 
from  the  sum,  if  there  are  3  required  numbers,  divide  by  3  for  the  smaller 
number;  if  4,  divide  by  4,  and  so  on. 

6.  The  product  of  two  numbers,  and  one  of  the  numbers 
being  given,  to  find  the  other  numbers.  —  Divide  the  product 
by  the  given  number  (Art.  62). 

7.  The  product  of  three  numbers,  and  two  of  the  numbers 
being  given,  to  find  the  other  number.  —  Divide  the  given  pro- 
duct by  the  product  of  the  two  given  numbers  (Art.  72). 

8.  The  dividend  and  quotient  being  given,  to  find  the  di- 
visor.—  Divide  the  dividend  by  the  quotient  (Art.  77). 

9.  The  divisor  and  quotient  being  given,  to  find  the  divi- 
dend. —  Multiply  the  divisor  and  quotient  together  (Art.  74). 


PROBLEMS.  67 


Examples. 


1.  A  carpenter  has  contracted  to  build  one  house  for  2763 
dollars,  another  for  4650  dollars,  and  a  third  for  8950  dollars. 
How  much  is  he  to  receive  for  them  all  ?     Ans.  1 6363  dollars. 

2.  N.  Chandler  has  invested  in  railroad  stock  and  a  small 
farm  929  dollars.  If  the  amount  invested  in  the  stock  was  279 
dollars,  how  much  did  the  farm  cost  him  ?    Ans.  650  dollars. 

3.  Mount  Black,  the  highest  peak  of  the  Blue  Ridge,  is  6476 
feet  high,  which  is  242  feet  higher  than  Mount  Washington,  the 
highest  peak  of  the  White  Mountains.  What  is  the  height  of 
Mount  Washington  ? 

4.  The  city  of  Mexico,  in  1519,  was  taken  by  Cortes,  and, 
328  years  after,  by  General  Scott.  In  what  year  did  it  yield 
to  Scott  ?  Ans.  1847. 

5.  Two  travellers,  A  and  B,  meeting  on  a  journey,  found 
they  had  both  travelled  1963  miles,  and  that  A  had  travelled 
199  miles  more  than  B.     What  distance  had  each  travelled  ? 

Ans.  A,  1081  miles ;  B,  882  miles. 

6.  A  father  gave  his  three  sons  4698  dollars,  of  which  James 
received  250  dollars  more  than  George,  and  Edwin  410  dollars 
more  than  George.     What  sum  did  each  receive  ? 

Ans.  George  $1346;  James  $1596;  Edwin  $1756. 

7.  There  was  paid  for  217  chests  of  tea  8463  dollars.  How 
much  was  that  a  chest  ? 

8.  How  many  weeks  will  684  bushels  of  oats  last  19  horses, 
each  horse  consuming  3  bushels  a  week  ? 

9.  On  dividing  3808  dollars  among  a  certain  number  of  men, 
it  was  found  that  the  share  of  each  was  224  dollars.  Eequired 
the  number  of  men.  Ans.  17  men. 

10.  A  certain  missionary  society  divides  its  income  among 
99  missions,  giving  to  each  an  average  of  575  dollars.  What 
is  its  income  ?  Ans.  56925  dollars. 

11.  The  product  of  96,  22,  and  one  other  number  is  63360. 
What  is  the  other  number  ?  Ans.  30. 

12.  The  divisor  being  13  and  the  quotient  1101,  what  is  the 
dividend?  Ans.  14313. 


68  MISCELLANEOUS   EXAMPLES. 


MISCELLANEOUS  EXAMPLES. 

1.  What  is  the  distance  by  railroad  from  Boston  to  Galena, 
it  being  from  Boston  to  Albany  200  miles,  from  Albany  to 
Niagara  Falls  305,  from  Niagara  Falls  to  Detroit  230,  from 
Detroit  to  Chicago  282,  from  Chicago  to  Galena  171  ? 

2.  Sold  J.  Weimer  my  best  horse  for  175  dollars,  my  second- 
best  chaise  for  87  dollars,  and  a  good  harness  for  31  dollars. 
He  has  paid  me  in  cash  38  dollars,  and  has  given  me  an  order 
on  S.  Lantz  for  12  dollars.     How  many  dollars  remain  due? 

3.  Bought  97  barrels  of  molasses  at  $  5  a  barrel.  Gave  17 
barrels  to  support  the  poor,  and  the  remainder  was  sold  at  $  8  a 
barrel.    Did  I  gain  or  lose,  and  how  much  ?    Ans.  $  155  gain. 

4.  It  requires  1728  cubic  inches  to  make  one  cubic  foot; 
required  the  number  of  cubic  inches  in  3787  cubic  feet. 

5.  If  a  garrison  of  987  men  are  supplied  with  175686  pounds 
of  beef,  how  much  will  there  be  for  each  man  ?    Ans.  178  lbs. 

6.  Albert  Peyton  sold  off  from  his  farm  120  acres,  gave  his 
son  80  acres,  and  had  remaining  1 60  acres ;  what  number  of 
acres  did  his  farm  contain  before  he  disposed  of  any  portion 
of  it?  Ans.  360  acres. 

7.  The  annual  revenue  of  a  gentleman  being  $  8395,  how 
much  per  day  is  that  equivalent  to,  there  being  365  days  in  a 
year?  Ans.  $23. 

8.  What  is  the  difference  between  half  a  dozen  dozen,  and 
six  dozen  dozen  ?  Ans.  792. 

9.  Bought  of  F.  Johnson  8  barrels  of  flour  at  $  7  per  barrel, 
and  3  hundred-weight  of  sugar  at  $  8  per  hundred.  What  was 
the  amount  of  his  bill  ? 

10.  George  Adams  bought  an  equal  number  of  cows  and 
oxen  for  3952  dollars.  For  the  cows  he  paid  31  dollars  each, 
and  for  the  oxen  45  dollars  each.  How  many  of  each  kind 
did  he  buy  ?  Ans.  52. 

11.  If  a  certain  quantity  of  provisions  will  sustain  13  men 
4  days,  how  long  would  it  sustain  1  man  ?     Ans.  52  days. 

12.  The  Globe  Manufacturing  Company  has  a  capital  of 
250,000  dollars,  divided  into  500  shares.  How  much  is  each 
share  ?  Ans.  500  dollars. 


MISCELLANEOUS   EXAMPLES.  69 

13.  Purchased  a  farm  of  500  acres  for  $17,876.  I  sold 
127  acres  of  it  at  $  47  an  acre,  212  acres  at  $  96  an  acre,  and 
the  remainder  at  $  37  an  acre.  What  did  I  gain  by  my  bar- 
gain ?  Ans.  $  14,402. 

14.  There  is  a  certain  island  18  miles  in  circuit,  which 
A  and  B  undertake  to  travel  round,  both  starting  from  the 
same  point  and  going  round  in  the  same  direction.  When 
A  has  travelled  17  miles  and  B  7  miles,  how  far  apart  are 
they  ?  Ans.  8  miles. 

15.  If  15  men  can  reap  a  certain  field  in  5  days,  how  long 
will  it  take  1  man  to  reap  the  same  ?  Ans.  75  days. 

16.  What  number  is  that,  which  being  multiplied  by  24, 
the  product  divided  by  10,  the  quotient  multiplied  by  2, 
32  subtracted  from  the  product,  the  remainder  divided  by 
4,  and  8  subtracted  from  the  quotient,  the  remainder  shall 
be  2  ?  Ans.  15. 

17.  From  126  +   (16  +  4)    X   2  take   (48   -4-   2)  + 


(34  X    6)   -r-   (17  —  5).  Ans.  125. 

18.  There  are  in  the  library  of  a  certain  school  683  books, 
which  number  will  give  23  books  to  each  pupil,  and  16  books 
over ;  what  is  the  number  of  pupils  ? 

19.  R.  Howland  in  making  a  journey,  after  walking  12 
miles  and  travelling  40  miles  by  stage,  went  by  steamboat  5 
times  the  distance  he  had  travelled  by  stage,  and  by  cars  6 
times  as  far  as  he  had  walked,  and  7  miles  besides.  What 
was  the  length  of  his  journey?  Ans.  331  miles. 

20.  At  an  election  A  and  B  were  candidates  for  the  same 
office,  and  the  whole  number  of  votes  cast  for  them  was  9891, 
of  which  B  received  1211  majority.  What  number  of  votes 
did  each  receive?  Ans.  A,  4340  ;  B,  5551. 

21.  The  product  of  3  numbers  is  4080 ;  one  of  the  numbers 
is  15,  and  another  16.     What  is  the  third  number  ? 

22.  How  many  tons  of  coal  at  6  dollars  a  ton  will  be  re- 
quired to  pay  for  17  yards  of  cloth  at  4  dollars  a  yard  and  32 
bushels  of  wheat  at  2  dollars  a  bushel  ?  Ans.  22  tons. 

23.  James  Cooper  has  manufactured  in  4  years  5608  pairs 
of  shoes,  making  each  successive  year  100  pairs  more  than 
the  year  before ;  how  many  pairs  did  he  manufacture  each 
year?  Ans.  1252;  1352;  1452;  1552. 


70 


UNITED   STATES   MONEY. 


24.  How  many  years  will  it  take  a  young  man  earning  45 
dollars,  and  spending  85  dollars  of  it,  every  month,  if  he  have 
already  620  dollars,  to  lay  up  enough  more  to  pay  for  a  house 
costing  1100  dollars?  Ans.  4  years. 

25.  If  the  product  of  the  divisor  by  the  quotient  be  19782, 
and  the  remainder  31,  what  is  the  dividend?    Ans.  19813. 

26.  John  Franklin  purchased  railroad  stock  to  the  amount 
of  4473  dollars,  and  sold  a  part  of  it  for  1885  dollars,  obtain- 
ing 65  dollars  a  share,  which  was  at  a  loss  of  6  dollars  on 
every  share  sold ;  but  the  stock  advancing  much  in  value,  he 
was  able  to  dispose  of  the  balance  so  as  to  gain  by  the  whole 
operation  812  dollars.  What  did  he  get  a  share  for  the  bal- 
ance of  the  stock  ?  Ans.  100  dollars. 


UNITED  STATES  MONEY. 

Art.  100t  United  States  Money  is  the  legal  currency 
of  the  United  States.  It  was  established  by  Congress  in  1786. 
Its  denominations  and  their  relative  values  are  shown  in  the 


following 

TABLE. 

10  Mills 

make 

1  Cent, 

marked 

ct. 

10  Cents 

u 

1  Dime, 

u 

d. 

10  Dimes 

u 

1  Dollar, 

a 

S- 

10  Dollars 

u 

1  Eagle, 

U 

E. 

Mills. 

Cents. 

10         = 

1 

Dimes. 

100          = 

10 

=                1 

Dollars. 

1000          = 

100 

=           10 

1 

Eagle. 

10000          = 

1000 

=         100 

10 

=           1 

In  accounts  and  ordinary  business  transactions,  the  only 
denominations  mentioned  are  dollars  and  cents,  eagles  being 
expressed  as  dollars,  dimes  as  cents,  and  mills  as  a  fraction  of 
a  cent. 

The  dollar  is  the  integer,  or  unit  of  measure  of  United 
States  money ;  and  therefore  dimes  are  tenths  of  a  dollar ; 
cents,  hundredths  ;  and  mills,  thousandths. 


UNITED    STATES   MONEY.  Yl 

101  •  Parts  of  a  dollar,  or  any  quantity  thus  expressed,  in 
tenths,  hundredths,  etc.,  are  termed  decimals,  or  fractions, 
whose  denominator  (Art.  68)  is  1,  with  one  or  more  ciphers 
annexed.  They  are  usually  expressed  by  simply  writing  the 
numerator  (Art.  68)  with  a  point  ( . )  before  it,  called  the  deci- 
mal point,  or  separatrix  ;  the  first  place  at  the  right  of  the 
point  being  tenths  ;  the  second  place,  hundredths  ;  the  third 
place,  thousandths ;  the  fourth  place,  ten-thousandths ;  and  so 
on.  Thus,  TV  is  written  .1 ;  TT^  is  written  .01 ;  To-W  *s  writ- 
ten .001  ;  ijjfau  is  written  .0001,  etc. 

102t  In  writing  dollars  and  cents  together,  the  decimal 
point  or  separatrix  is  placed  between  the  dollars  and  the  cents 
or  decimal  part ;  and  since  cents  occupy  two  places,  the  place 
of  dimes  and  of  cents,  when  the  number  of  cents  is  less  than 
10,  a  cipher  must  be  written  before  them,  in  the  place  of  dimes. 
Thus,  $30,375  is  read,  thirty  dollars  thirty-seven  cents  five 
mills,  or  thirty  dollars  and  three  hundred  seventy-five  thou- 
sandths of  a  dollar ;  $  12.05  is  read,  twelve  dollars  five  cents, 
or  twelve  dollars  and  five  hundredths  of  a  dollar,  etc. 

103 1  The  denominations  of  United  States  Money  increasing 
from  right  to  left,  and  decreasing  from  left  to  right,  in  the  same 
manner  as  do  the  units  of  the  several  orders,  as  simple  whole 
numbers,  they  may  therefore  be  added,  subtracted,  multiplied, 
and  divided  according  to  the  same  rules. 

104.  The  coins  of  the  United  States  are  of  gold,  silver, 
and  copper. 

The  gold  coins  are  the  double  eagle,  eagle,  half-eagle,  quar- 
ter-eagle, three-dollars,  and  dollar. 

The  silver  coins  are  the  dollar,  half-dollar,  quarter-dollar, 
dime,  half-dime,  and  three-cent-piece. 

The  copper  coins  are  the  cent  and  half-cent. 

Note.  —  All  the  gold  and  silver  coins  of  the  United  States  are  now  made  of 
one  purity,  9  parts  of  pure  metal  and  1  part  alloy.  The  alloy  for  the  silver  is 
copper,  and  that  for  the  gold  1  part  copper  and  1  part  silver.  The  cent  and 
half-cent,  old  coinage,  are  of  pure  copper;  the  cent,  new  coinage,  is  made 
88  parts  copper  and  12  parts  nickel. 

The  standard  weight  of  the  eagle,  as  fixed  by  present  laws,  is  258  grains 
Troy,  and  the  gold  coins  in  proportion  according  to  their  values.  The  weight 
of  the  silver  dollar  is  412 \  grains;  half-dollar,  192  grains;  quarter-dollar,  9G 


72  UNITED   STATES   MONEY. 

grains ;  dime  38|  grains ;  half-dime,  19|  grains ;  three-cent  piece,  11^  grains ; 
the  cent,  old  coinage,  168  grains;  the  cent,  new  coinage,  72  grains. 

The  weight  of  the  silver  dollar,  it  will  be  seen,  is  greater  in  proportion  to  its 
value  than  the  other  silver  coins.  This  is  owing  to  their  standard  weight  hav- 
ing been  reduced,  while  that  of  the  dollar  remained  unchanged ;  but  since  this 
reduction  of  weight  of  the  smaller  silver  coins,  no  more  of  the  silver  dollar  ap- 
pear to  have  been  coined.  In  circulation,  the  gold  dollar,  which  of  late  years 
has  been  extensively  coined,  has  almost  entirely  taken  the  place  of  the  silver 
dollar. 

The  symbol  $,  or  dollar  sign,  represents,  probably,  the  letter  U  written 
upon  an  S,  denoting  U.  S.  (United  States). 


REDUCTION  OF  UNITED   STATES  MONEY. 

105.  Reduction  of  United  States  Money  is  changing  the 
units  of  one  of  its  denominations  to  the  units  of  another,  either 
of  a  higher  or  lower  denomination,  without  altering  their  value. 

106.  To  reduce  units  from  a  higher  denomination  to  a 
lower. 

Ex.  1.     Reduce  58  dollars  to  cents  and  mills. 

Ans.  5800  cents ;  58000  mills. 

OPERATION. 

5  8  dollars. 

\  0  q  Wc  multiply  the  58  dollars  by 

100  to  reduce  them  to  cents,  be- 

5  8  00  cents.  cause  100  cents  make  1  dollar; 

1  0  and  multiply  the  cents  by  10  to 

reduce  them  to  mills,  because  10 

5  8  0  0  0  mills  make  1  cent.     Hence, 

Or  thus :  5  8  0  0  0  mills. 

To  reduce  dollars  to  cents,  annex  two  ciphers  ;  to  reduce  dollars  to 
mills,  annex  three  ciphers;  and  to  reduce  cents  to  mills,  annex  one 
cipher. 

Note.  —  Dollars,  cents,  and  mills,  expressed  by  a  single  number,  are  re- 
duced to  mills  by  merely  removing  the  separating  point;  and  dollars  and 
cents,  by  annexing  one  cipher  and  removing  the  separatrix. 

107.  To  reduce  units  from  a  lower  denomination  to  a  higher. 
Ex.  1.     Reduce  58000  mills  to  cents  and  to  dollars. 

Ans.  5800  cents;  58  dollars. 

operation.  "We  divide  the  mills  by  10  to  re- 

10)  58000  mills.        duce  them  to  cents,  because  10  mills 

i  a  a  \  k  a  a  a         *  make  1  cent;  and  the  cents  by  100 

1  0  0  )5800  cents.  to  reduce  th'em  to  dollarg)  ^cause 

5  8  dollars.  100  cents  make  1  dollar.     Hence, 


UNITED    STATES   MONEY.  73 

To  reduce  mills  to  cents,  point  off  one  figure  on  the  right ;  to  reduce 
cents  to  dollars,  point  off  two  figures ;  and  to  reduce  mills  to  dollars, 
point  off  three  figures. 

Examples. 

2.  Reduce  $765  to  cents. 

3.  Change  726  mills  to  cents.  Ans.  72^  cents. 

4.  How  many  dollars  are  329  cents?  Ans.  $3.29. 

5.  Change  12345  mills  to  dollars.  Ans.  $  12.345. 

6.  Reduce  $  123.56  to  mills.  Ans.  123560  mills. 

7.  Reduce  2  eagles,  2  dollars,  and  2  dimes  to  cents. 

Ans.  2220  cts. 

ADDITION  OF  UNITED   STATES  MONEY. 

108.  Ex.  1.  Add  together  17  dollars  13  cents  5  mills ; 
8  dollars  4  cents  6  mills ;  63  dollars  20  cents  3  mills ;  and 
29  dollars  87  cents  5  mills.  Ans.  $  118.259. 


We  write  units  of  the  same  denomina- 
tion in  the  same  column,  and  add  as  in 
addition  of  simple  numbers  (Art.  45),  and 
separate  the  dollars  from  the  cents  in  the 
answer  by  the  decimal  point. 


OPERATION. 

$     cts.  m. 

1  7.1  3  5 

8.0  4  6 

6  3.2  0  3 

.29.875 

Ans.  1  1  8.2  5  9 


Rule.  —  Write  dollars,  cents,  and  mills,  so  that  units  of  the  same 
denomination  shall  stand  in  the  same  column. 

Add  as  in  addition  of  simple  numbers,  and  place  the  decimal  point 
directly  under  that  above. 

Proof.  —  The  proof  is  the  same  as  in  addition  of  simple 
numbers. 

Examples. 

2.                        3.                         4.  5. 

$       cts.  m.                  $     cts.  m.                    $       cts.  m.  $     cts.  m. 

3  7  5.875     78.193     171.013  861.073 

6  7  1.1  2  7     1  8.0  1  4     3  8  2.0  9  4  5  1  6.7  1  6 

3  8  7.143     91.038     999.900  344.673 

1  8  4.1  8  9     1  6.8  1  7     1  5  5.0  6  8  6  1  7.8  1  4 

14  7.758     81.476      48.153  169.973 

6  3.0  7  2     4  3.18  4      4  9.6  19  8  10.426 


18  29.16  4 


N  7 


74  UNITED   STATES   MONEY. 

6.  Add  the  following  sums,  $18,165,  $701.63,  $151,161, 
$  375.089,  and  $  471.017.  Ans.  $  1717.062. 

7.  Bought  a  horse  for  eighty-seven  dollars  nine  cents,  a  pair 
of  oxen  for  sixty-five  dollars  twenty  cents,  and  six  gallons  of 
molasses  for  two  dollars  six  cents  five  mills ;  what  was  the 
amount  of  my  bill  ?  Ans.  $  154.355. 

8.  Sold  a  calf  for  three  dollars  eight  cents,  a  bushel  of  corn 
for  ninety-seven  cents  five  mills,  and  three  bushels  of  rye  for 
three  dollars  five  cents ;  what  was  the  amount  received  ? 

Ans.  $  7.105. 

SUBTRACTION  OF  UNITED   STATES  MONEY. 

109.     Ex.  1.  From  106  dollars  7  cents  7  mills,  take  92 

dollars  83  cents  8  mills.  Ans.  $  13.239. 

operation.  "We  write   the   less    number   under   the 

$     cts.  m.  greater,  place  mills  under  mills,  cents  under 

1  0  6.0  7  7  cents,  and  dollars  under  dollars,  and  subtract 

9  2.8  3  8  as  in  subtraction  of  simple  numbers  (Art. 

a            o0qq  50),  and  separate  the  dollars  from  the  cents 

Ans.  1  6,1  6  J  jn  ^e  answer  by  the  decimal  point. 

Rule.  —  Write  the  several  denominations  of  the  subtrahend  under 
the  corresponding  ones  of  the  minuend. 

Subtract  as  in  subtraction  of  simple  numbers,  and  place  the  decimal 
point  directly  under  that  above. 

Proof,  —  The  proof  is  the  same  as  in  subtraction  of  simple 
numbers. 

Examples. 

2.  3.  4.  5. 


|     cts.  m. 

8  7  1.1  6  1 
8  9.9  1  8 

|     cts.  m. 

4  7  8.4  7  7 
1  9  9.9  9  1 

$      cts.  m. 

1  6  7.1  6  3 

9  8.0  9  7 

$      cts.  m. 

1  6  3.1  6  7 

9.0  9  8 

7  8  1.2  4  3 

6.  Bought  a  farm  for  $1728.90,  and  sold  it  for  $3786.98; 
what  did  I  gain  by  my  bargain  ? 

7.  Gave  $79.25  for  a  horse,  and  $106,875  for  a  chaise,  and 
sold  them  both  for  $  200 ;  what  did  I  gain  ? 

8.  Bought  a  farm  for  $8967,  and  sold  it  for  nine  thousand 
eight  hundred  seventy-six  dollars  seventy-five  cents ;  what  did 
I  gain  ?  Ans.  $  909.75. 


UNITED   STATES   MONEY.  75 

9.  Bought  a  barrel  of  flour  for  $  7.50,  three  bushels  of  rye 
for  $  2.75,  and  three  cords  of  wood  at  $  5.25  a  cord  ;  I  sold  the 
flour  for  $  6.18,  the  rye  for  $  3.00,  and  the  wood  for  $  6.75  a 
cord ;  what  was  gained  by  the  bargain  ?  Ans.  $  3.43. 

10.  A  young  lady  went  a  "  shopping."  Her  father  gave  her 
a  twenty-dollar  bill.  She  purchased  a  dress  for  $  8.16,  a  muff 
for  $  3.19,  a  pair  of  gloves  for  $  1.12,  a  pair  of  shoes  for  $  1.90, 
a  fan  for  $  0.19,  and  a  bonnet  for  $  3.08 ;  how  much  money  did 
she  return  to  her  father  ?  Ans.  $  2.36. 

MULTIPLICATION  OF  UNITED   STATES  MONEY. 

110.  Ex.  1.  What  will  365  barrels  of  flour  cost  at  $5.75 
a  barrel  ?  Ans.  $  2098.75. 

OPERATION. 

$  5.7  5  We  multiply  as  in  multiplication  of  sim- 

3  6  5  pie  numbers,  and,  obtaining  the  product  in 

o  o  7  k  cents,  the  lowest  denomination  of  the  multi- 

o  a  n  n  plicand,  we  reduce  them  to  dollars  by  point- 

^  4  „  v  ing  off  two  places  at  the  right  for  cents 

1725  (Art.  107). 


$2  0  9  8.7  5  Ans. 

Rule.  —  Multiply  as  in  multiplication  of  simple  numbers. 
The  product  will  be  in  the  lowest  denomination  in  the  multiplicand, 
which  must  be  pointed  off  as  in  reduction  of  United  States  money. 

Proof.  —  The  proof  is  the  same  as  in  multiplication  of  simple 
numbers. 

Examples. 

2.  What  will  126  pounds  of  butter  cost  at  13  cents  a  pound? 

3.  What  will  63  pounds  of  tea  cost  at  93  cents  a  pound  ? 

4.  What  will  43  tons  of  hay  cost  at  13  dollars  75  cents  a 
ton?  Ans.  $591.25. 

5.  If  1  pound  of  pork  is  worth  7  cents  3  mills,  what  are  46 
pounds  worth?  Ans.  $3,358. 

6.  If  1  hundred  of  beef  cost  3  dollars  28  cents,  what  are  76 
hundred  worth  ? 

7.  What  will  96  thousand  feet  of  boards  cost  at  11  dollars 
67  cents  a  thousand?  Ans.  $  1120.32. 

8.  If  a  barrel  of  cider  be  sold  for  2  dollars  12  cents,  what  will 
be  the  value  of  169  barrels  ?  Ans.  $  358.28. 


76  UNITED   STATES   MONEY. 

9.  What  will  be  the  value  of  a  hogshead  of  wine,  containing 
63  gallons,  at  1  dollar  63  cents  a  gallon?         Ans.  $  102.69. 

10.  Sold  a  sack  of  hops,  weighing  396  pounds,  at  11  cents  3 
mills  a  pound ;  to  what  did  it  amount  ?  Ans.  $  44.748. 

11.  Sold  19  cords  of  wood,  at  $5.75  a  cord;  to  what  did  it 
amount  ? 

12.  Sold  169  tons  of  timber,  at  $4.68  a  ton;  what  did  I 
receive  ? 

13.  Sold  a  hogshead  of  sugar,  weighing  465  pounds ;  to  what 
did  it  amount,  at  14  cents  a  pound?  Ans.  $  65.10. 

14.  What  will  be  the  amount  of  789  pounds  of  leather  at  18 
cents  a  pound?  Ans.  $  142.02. 

15.  What  will  be  the  expense  of  846  pounds  of  sheet-lead  at 
5  cents  7  mills  a  pound  ?  Ans.  $  48.222. 

16.  When  potash  is  sold  for  $  132.55  a  ton,  what  will  be  the 
price  of  369  tons  ?  Ans.  $  48910.95. 

17.  What  will  365  pounds  of  beeswax  cost  at  18  cents  4 
mills  a  pound  ?  Ans.  $  67.16. 

18.  If  1  pound  of  tallow  cost  7  cents  3  mills,  what  are  968 
pounds  worth  ?  Ans.  $  70.664. 

DIVISION  OF  UNITED   STATES  MONEY. 

111.     Ex.  1.     If  78  barrels  of  fish  cost  $  303.42,  what  will 
1  barrel  cost  ?  Ans.  $  3.89. 

OPERATION. 

78)  $3  0  3.4  2  (  $3.8  9 

2  3  4  We  divide  as  in  division  of  sim- 

nQ  *  ple  numbers,  and  point  off  as  many 

places  for  cents  in  the  quotient  as 

6  2  4  there  are  for  cents   in  the    divi- 

7  0  2  dend- 

702 


Rule.  —  Divide  as  in  division  of  simple  numbers. 
The  quotient  will  be  in  the  lowest  denomination  of  the  dividend,  which 
must  be  pointed  off  as  in  reduction  of  United  States  money. 

Note.  —  When  the  dividend  consists  of  dollars  only,  and  is  either  smaller 
than  the  divisor  or  not  divisible  by  it  without  a  remainder,  reduce  it  to  a  lower 
denomination  by  annexing  two  or  three  ciphers,  as  the  case  may  require,  and 
the  quotient  will  be  cents  or  mills  accordingly. 

When  it  is  required  to  find  the  number  of  times  one  sum  of  money  is  con- 
tained in  another,  both  sums,  if  of  different  denominations,  must  be  reduced  to 
the  same,  before  dividing. 


3.    How    many   yards   of 
cloth  at  9  cents  a  yard  can 
be  bought  for  $  28.89  ? 

Ans.  321  yards. 

OPERATION. 

9)2889 

Ans.       3  2  1  yards. 

UNITED   STATES   MONEY.  77 

Proof.  —  The  proof  is  the  same  as   in  division  of  simple 
numbers. 

Examples. 

2.  Bought  11  chests  of  tea 
for  $234.30.  What  did  I  give 
a  chest  ?         Ans.  $  21.30. 

OPERATION. 

11)  $2  3  4.3  0 
Ans.    $2  1.3  0 

4.  When  19  bushels  of  fine  salt  can  be  bought  for  $30.87£, 
what  costs  one  bushel  ? 

5.  At  12  cents  a  pound,  how  many  pounds  of  sugar  can  be 
bought  for  $51? 

6.  If  78  barrels  of  fish  cost  $303.42,  what  will  one  barrel 
cost? 

7.  Bought  42  bushels  of  pears  for  $  73.50  ;  what  cost  one 
bushel?  Ans.  $1.75. 

8.  How  many  yards  of  broadcloth  at  $  2.75  a  yard  can  be 
bought  for  $  904.75  ?  Ans.  329  yards. 

9.  When  rye  is  sold  at  the  rate  of  628  bushels  for  $  471.00, 
what  is  that  a  bushel  ?  Ans.  $  0.75. 

10.  If  it  should  cost  $1460  to  support  a  family  365  days, 
what  would  be  the  expense  a  day  ?  Ans.  $  4. 

11.  How  many  gallons  of  oil  at  $  1.62 J-  a  gallon  can  be 
bought  for  $  234  ? 

12.  If  1624  pounds  of  pork  cost  $97.44,  what  cost  one 
pound?  Ans.  $0.06. 

13.  If  47  thousand  shingles  cost  $  176.25,  what  is  the  cost 
per  thousand  ?  Ans.  $  3.75. 

14.  Bought  148  tons  of  plaster  of  Paris  for  $  337.44;  what 
was  it  per  ton  ?  Ans.  $  2.28. 

15.  At  $37.75  an  acre,  how  many  acres  of  land  may  be 
bought  for  $  1774.25  ?  Ans.  47  acres. 

16.  At  67  cents  a  pound,  how  many  chests  of  tea,  each  weigh- 
ing 59  pounds,  can  be  bought  for  $  672.01  ?    Ans.  17  chests. 

17.  How  many  bushels  of  wheat  at  $  1.25  a  bushel  can  be 
bought  for  $  863.75  ?  Ans.  691  bushels. 

7* 


78  UNITED   STATES   MONEY. 

18.  Sold  169  tons  of  timber  for  $790.92;  what  cost  one 
ton? 

19.  "What  cost  one  pound  of  leather,  if  789  pounds  cost 
$  142.02  ? 

20.  If  369  tons  of  potash  cost  $48910.95,  what  will  be  the 
price  of  one  ton  ?  Ans.  $  132.55. 

21.  Bought  47  hogsheads  of  salt,  each  hogshead  containing 
7  bushels,  for  $  368.48 ;  what  cost  one  bushel  ?    Ans.  $  1.12. 

GENEKAL  PRINCIPLES  AND  APPLICATIONS. 

112.  Analysis  is  an  examination  of  a  question  by  re- 
solving it  into  its  parts,  in  order  to  consider  them  separately, 
and  thus,  by  reasoning  from  the  nature  of  the  question,  to 
render  each  step  in  the  solution  plain  and  intelligible. 

113.  An  aliquot  part  of  any  number  or  quantity  is  such  a 
part  as  will  exactly  divide  that  number  or  quantity.  Thus,  3 
is  an  aliquot  part  of  6 ;  and  4,  an  aliquot  part  of  16. 

114.  The  method  of  performing  operations  by  means  of 
aliquot  parts  is  called  Practice,  or  Analysis  by  Aliquot 
Parts. 

115.  In  business  calculations  frequent  use  is  made  of  the 
following 

Aliquot  Parts  of  a  Dollar. 


50  cents  =  £  of  $1. 
33£  cents  =  £  of  $  1. 
25  cents  =  £  of  $1. 
20    cents  =  4  of  $  I . 


12i  cents  =   £  of  $  1, 
10    cents  =  T\j-  of  $  1. 

6£  cents  =  T\  of  $  1. 

5    cents  =  J$  of  $  1. 


Practical  Questions  by  Analysis. 

116.  The  price  of  a  unit  of  any  quantity  being  given,  to 
find  the  cost  of  that  quantity. 

Ex.  1.  If  1  barrel  of  flour  cost  $7,  what  will  125  barrels 
cost? 

Analysis.  Since  1  barrel  costs  $  7,  125  barrels  will  cost 
125  times  as  much  :  $  7  X  125  =  $  875. 


GENERAL   PRINCIPLES   AND   APPLICATIONS.  79 

2.  If  1  pound  of  beef  cost  13  cents,  what  will  914  pounds 
cost?  Ans.  $118.82. 

3.  If  1  yard  of  calico  cost  23  cents,  what  will  31L  yards 
cost  ? 

4.  When  44  cents  are  paid  for  1  pound  of  tea,  how  much 
must  be  paid  for  15  chests  of  tea,  each  containing  47  pounds  ? 

5.  At  39 J  cents  a  pound,  how  much  must  be  paid  for  9 
bales  of  wool,  each  containing  317  pounds  ?   Ans.  $  1126.93J. 

6.  If  the  cost  of  building  1  mile  of  railroad  be  $41315, 
what  will  be  that  of  113  miles  ?  Ans.  $  4668595. 

117#  The  price  of  a  unit  of  any  quantity  being  a  given 
aliquot  part  of  a  dollar,  to  find  the  cost  of  that  quantity. 

7.  At  12£  cents  a  yard,  what  will  208  yards  of  calico 
cost?  Ans.  $26. 

Analysis.  Since,  at  $  1  a  yard,  the  cost  would  be  as  many 
dollars  as  there  are  yards,  the  cost  at  12  J  cents,  or  J  of  $  1, 
will  be  one  eighth  as  many  dollars  as  there  are  yards,  or  as 
many  dollars  as  8  is  contained  times  in  208  ;  208  -7-8  =  26. 
Therefore  208  yards  cost  $26.  That  is,  we  take  such  a 
part  of  the  number  denoting  the  quantity  as  the  price  is  of  1 
dollar, 

8.  What  will  be  the  cost  of  362  pounds  of  feathers  at  33£ 
cents  a  pound  ?  Ans.  $  120.66§. 

9.  At  50  cents  a  bushel,  what  will  7  loads  of  potatoes  cost, 
each  containing  30  bushels  ?  Ans.  $  105. 

10.  At  6£  cents  a  pound,  how  much  must  be  paid  for  1163 
pounds  of  shingle  nails  ? 

11.  What  will  be  the  cost  of  96  yards  of  broadcloth  at 
$4.33^  a  yard?  Ans.  $416. 

Analysis.  Since,  at  $  1  a  yard,  the  cost  would  be  as  many 
dollars  as  there  are  yards,  the  cost  at  $  4.33^,  or  $  4£,  will 
be  4£  times  as  much  :  $  4£  X  96  =  $  416. 

12.  If  the  cost  of  1  pair  of  boots  be  $3.16|,  what  will  be 
the  cost  of  20  cases,  each  case  containing  60  pairs  ? 

13.  At  $  2.25  each,  what  will  150  hats  cost  ? 

Ans.  $  337.50. 

14.  What  will  98  dozen  knives  cost  at  $  5.12^  a  dozen  ? 

Ans.  $  502.50. 


80  UNITED   STATES  MONEY. 

15.  If  James  Spooner  spends  uselessly  every  clay  for  cigars 
G£  cents,  how  much  does  he  thus  spend  in  360  days  ? 

16.  At  $  2.20  a  bushel,  what  will  15  car-loads  of  wheat,  each 
containing  212  bushels,  cost  ?  Ans.  $  6996. 

118.     The  price  of  any  article  sold  by  the  100,  or  1000, 
with  the  quantity,  being  given,  to  find  the  cost  of  that  quantity. 

17.  What  will  732  fishes  cost  at  $  5  a  hundred? 

Ans.  $36.60. 
Analysis.  Since  100  fishes  cost  $  5,  1  fish  will  cost  -j-^  of 
$  5,  or  $  Tfo  ;  and  if  1  fish  cost  $  Tfo,  732  fishes  will  cost  732 
times  as  much:  $  5  X  732  =  $3660;  $3660  ~-  100  = 
$36.60.  That  is,  we  multiply  the  quantity  and  price  together ; 
and  cut  off  two  figures  at  the  right  (Art.  79). 

18.  How  much  must  be  paid  for  950  apple-trees  at  20  dol- 
lars a  hundred  ? 

19.  What  will  7235  arithmetics  cost  at  $  45  a  hundred  ? 

Ans.  $  3255.75. 

20.  What  will  it  cost  to  excavate  19875  solid  feet  of  earth,  at 
the  rate  of  52  cents  per  hundred  solid  feet?     Ans.  $  103.35. 

21.  What  is  the  value  of  1765  feet  of  timber  at  $  3  a  hun- 
dred? Ans.  $52.95. 

22.  What  is  the  value  of  2355  cedar  rails  at  $5.50  per 
hundred  ? 

23.  What  will  3100  bricks  cost  at  $  6  a  thousand? 

Ans.  $  18.60. 
Analysis.     Since  1000  bricks  cost  $  6,  1  brick  will  cost 

tttW  of  $  6>  or  $  tcjW  5  and>  if  *  brick  cost  $  tthttd  3*00  wm 
cost  3100  times  as  much :  $  6  X  3100  =  $  18600 ;  $  18600  ~- 
1000  =  $  18.60.  That  is,  we  multi-ply  the  quantity  and  price 
together,  and  cut  off  three  figures  at  the  right  (Art.  79). 

24.  At  $  142.50  a  thousand,  what  will  6150  chestnut  posts 
cost?  Ans.  $876,375. 

25.  How  much  must  be  paid  for  15750  feet  of  boards  at 
$  30  a  thousand  ? 

26.  At  $  2.25  per  thousand,  how  much   must  be  paid  for 
3550  laths  ?  Ans.  $  7.98*. 

27.  I  have  bought  of  L.  T.  Bobbins  363  feet  of  plank  at 


GENERAL   PRINCIPLES   AND   APPLICATIONS.  81 

$  2.50  per  hundred  ;  3150  pickets  at  $  14  per  thousand  ;  and 
350  feet  of  boards  at  $  20  per  thousand.  What  is  the  amount 
of  the  whole  ?  Ans.  $  60.17£. 

28.  What  must  be  paid  for  92350  railroad  ties,  at  $  135  per 
thousand  ? 

119.  The  price  of  any  article  sold  by  the  ton  of  2000 
pounds,  with  the  quantity,  being  given,  to  find  the  cost  of  that 
quantity. 

29.  At  $  8  a  ton,  what  will  2550  pounds  of  coal  cost  ? 

Ans.  $  10.20. 
Analysis.  Since  2000  pounds  cost  $  8,  1000  pounds  will 
cost  one  half  of  $  8,  or  $  4 ;  and,  if  1000  pounds  cost  $  4,  1 
pound  will  cost  TxjW  °f  $  4,  or  $  yxny^,  and  2550  pounds  will 
cost  2550  times  as  much :  $  4  X  2550  =  $  10200 ;  $  10200  ~- 
1000  ==  $  10.20.  That  is,  we  multiply  the  number  denoting 
the  quantity  and  half  the  price  together \  and  cut  off  three  figures 
at  the  right  (Art.  79). 

30.  How  much  must  be  paid  for  3760  pounds  of  hay,  when 
the  price  is  $  15  a  ton  ?  Ans.  $  28.20. 

31.  What  is  the  value  of  19  casks  of  plaster  of  Paris,  each 
weighing  500  pounds,  at  $  9  a  ton  ? 

32.  How  much  must  be  paid  for  transporting  5163  pounds 
of  freight,  at  $  3.50  per  ton  ? 

33.  At  $  46  a  ton,  what  will  33550  pounds  of  railroad  iron 
cost?  ,%  Ans.  $771.65. 

34.  G.  Reed  has  bought  30  bags  of  superphosphate  of  lime, 
each  bag  containing  125  pounds,  at  $45  per  ton;  and  7500 
pounds  of  Peruvian  guano,  at  $  53  per  ton.  What  is  the  cost 
of  the  whole  ?  Ans.  $  283.125. 

35.  I  have  bought  6350  pounds  of  anthracite  coal  at  $7.12J 
per  ton,  and  3560  pounds  of  Pictou  coal  at  $7.25  per  ton. 
What  is  the  cost  of  the  whole  ?  Ans.  $  35.526J. 

120*  The  price  of  any  quantity,  and  the  quantity,  being 
given,  to  find  the  price  of  a  unit  of  that  quantity. 

36.  If  125  barrels  of  flour  cost  $875,  what  will  1  barrel 
cost  ?  Ans.  $  7. 

Analysis.     Since  125  barrels  cost  $875,  1  barrel  will  cost 


82  UNITED   STATES   MONEY. 

as  many  dollars  as  125  is  contained  times  in  875  ;  $  875  -~ 

125  =  $7. 

37.  If  914  pounds  of  beef  cost  $  118.82,  what  will  1  pound 
cost?  Ans.  $0.13. 

38.  If  96  yards  of  broadcloth  can  be  bought  for  $400,  for 
what  can  1  yard  be  bought  ?  Ans.  $  4.1 6f. 

39.  If  $  510  are  paid  for  120  pairs  of  boots,  what  is  the  cost 
of  1  pair  ?     What  is  the  cost  of  17  pairs  ? 

40.  A  merchant  paid  $  270  for  300  casks  of  lime ;  what 
did  he  pay  a  cask  ?     What  did  he  pay  for  33  casks  ? 

Ans.  $  29.70. 

41.  When  boards  are  $25  per  thousand,  what  are  they  per 
foot  ?     What  per  hundred  ?  Ans.  $  2.50. 

42.  If  3  pieces  of  calico,  each  containing  30  yards,  cost 
$  19.80,  what  does  1  yard  cost  ?  Ans.  $  0.22. 

43.  When  10  firkins  of  butter,  each  containing  5G  pounds, 
can  be  bought  for  $  112,  what  is  the  price  per  pound  ? 

Ans.  $0.20. 
121.  The  price  of  any  quantity,  and  the  price  of  a  unit  of 
that  quantity,  being  given,  to  find  the  quantity. 

44.  At  $  7  per  barrel,  how  many  barrels  of  flour  can  be 
purchased  for  $875  ?  Ans.  125  barrels. 

Analysis.  Since  1  barrel  will  cost  $  7,  as  many  barrels  can 
be  purchased  with  $  875  as  7  i3  contained  times  in  875 ;  875 
-f-  7  =  125  ;  therefore,  there  can  be  purchased  125  barrels. 

45.  At  $  3  a  yard,  how  many  yards  of  broadcloth  can  I 
purchase  with  $456  ?  Ans.  152  yards. 

46.  How  many  cords  of  wood,  at  $  4.25  each,  can  be  bought 
for  $  357  ? 

47.  At  33^-  cents  a  pound,  how  many  pounds  of  feathers  can 
be  bought  for  $  120.66§?  Ans.  362  pounds. 

48.  At  50  cents  a  bushel,  how  many  barrels  of  apples,  each 
containing  3  bushels,  can  be  purchased  for  $  40.50  ? 

Ans.  27  barrels. 

49.  Expended  for  hemp  $11600,  at  $232  per  ton.  How 
many  tons  did  I  buy  ?  Ans.  50  tons. 

50.  How  many  boxes  of  raisins,  at  $  3.50  per  box,  can  be 
bought  for  $  756  ?  Ans.  216  boxes. 


BILLS.  83 


BILLS. 


122.  A  Bill  is  a  paper,  given  by  merchants,  containing  a 
statement  of  goods  sold,  and  their  prices. 

An  invoice  is  a  bill  of  merchandise  shipped  or  forwarded  to 
a  purchaser,  or  selling  agent. 

The  date  of  a  bill  is  the  time  of  the  transaction. 

The  bill  is  against  the  party  owing,  and  in  favor  of  the 
party  who  is  to  receive  the  amount  due. 

A  bill  is  receipted,  when  the  receiving  of  the  amount  due  is 
acknowledged  by  the  party  in  whose  favor  it  is,  as  in  bill  1. 
A  clerk,  or  any  other  authorized  person,  may,  in  his  stead, 
receipt  for  him,  as  in  bill  2. 

When  the  items  of  a  bill  have  been  rendered  at  different 
dates,  the  several  times  may  be  given  at  the  left  hand,  as  in 
bill  4. 

When  the  bill  is  in  the  form  of  an  account,  containing  items 
of  debt  and  credit  in  its  settlement,  it  is  required  to  find  the 
difference  due,  or  balance,  as  in  bill  4. 

Note.  —  A  due-bill  is  a  ■written  acknowledgment  that  a  debt  is  due,  and  is 
often  given  in  making  settlements  to  remove  cause  of  subsequent  disputes  as 
to  a  claim.    A  bill  is  sometimes  settled  or  receipted  by  a  due-bill,  as  in  bill  4. 

What  is  the  cost  of  each  article  in,  and  the  amount  due  of, 
each  of  the  following  bills  ? 


(1-) 

Be 

)ston,  July  4,  ! 

Mr*  James  Dow, 

Bought  of 

Dennis  Shai 

17 

yds.  Flannel, 

at 

$0.45 

19 

"     Shalloon, 

a 

.37 

16 

"     Blue  Camlet, 

a 

.46 

13 

"     Silk  Vesting, 

a 

.87 

9 

"     Cambric  Muslin, 

a 

.63 

25 

"     Bombazine, 

a 

.56 

17 

"      Ticking, 

ft 

.31 

19 

"     Striped  Jean, 
Received  payment, 

u 

.16 

']? 

Dennis  Sharp. 

61.33 


84 


UNITED   STATES   MONEY. 


(2.) 

Chicago,  Jan.  1,  1857. 

•.  Samuel  Smith, 

Bought  of 

David  Johnson, 

13  lbs.  Tea,                          at 

$0.98 

16    "    Coffee, 

.15 

36    "    Sugar,                        " 

.13 

47    "    Cheese,                       " 

.09 

12    "    Pepper,                    '.« 

.19 

7    "    Ginger,                *•     " 

.17 

13    "    Chocolate,                 " 

.61 

$ 

Received  payment, 

David  Johnson, 

by  J.  Keen 

(3.)  Philadelphia,  Sept  5,  1856. 

Messrs.  Wilson,  Niles,  &  Co., 

Bought  of  Peck  &  Bliss, 
2  doz.  National  Arithmetics,  at  %  6.00 


3    "     Coin.  Sch.  Arithmetics,              " 

5.00 

5    "     Primary  Arithmetics,                 " 
17  Parker's  Exercises  in  Composition,  " 

L80 
.25 

13  Maglathliris  National  Speaker,         " 
19  Lever etC s  Ccesar,                                 " 

.60 
.50 

3  Greenleafs  Algebra,                          " 
7  Sillimaris  Chemistry,                        " 
15  6V?^s  £71  aSI  History,                       " 
5  imp.  Quarto  Bibles,                          " 
3  Webster's  Quarto  Dictionaries,         " 
5  Felton's  Greek  Readers,                    " 

.60 
1.12* 

.22 

15.00 

4.50 

1.50 

1  Kane's  Arctic  Explorations,              " 

4.50 

<2  171 

Received  payment, 

Peck  &  Bliss. 

BILLS. 


85 


(4.)  New  Orleans,  Jan.  1,  1856. 

Mr.  Albert  Crawford, 

To  Baldwin,  Sherwood,  &  Bice,  Dr. 
1855. 


Jan.    2. 

.For  17  yefo. 

,  Broadcloth, 

at 

$5.25 

"     15. 

"     29    " 

Gassimere, 

u 

1.62 

Feb.    3. 

"     60    " 

Bleached  Shirting, 

a 

.17 

"•      7. 

u     49    « 

Ticking, 

a 

.27 

"     15. 

«     18    " 

Blue  Cloth, 

a 

3.19 

June  17. 

«     27    " 

Habit  do. 

a 

2.75 

Aug.   3. 

"    75    " 

Flannel, 

u 

.61 

Sept.  19. 

"    36    " 

Plaid  Prints, 

a 

.75 

Dec.    2. 

u    49    « 

Brown  Sheeting, 

a 

•      .18 

4b  Q70  on 

Cr. 

Jan.  28. 

i?^  cash, 

$83.00 

Feb.  19. 

"   3  if.  ^oar^, 

at 

$  30.00 

May    9. 

"   7  M.  > 

Shingles, 

a 

4.00 

June  17. 

"   4  efa^s'  Zafor, 

u 

2.00 

Ji%    3. 

"  5     u 

a 

a 

1.75 

Oct.    7. 

"   7  C.  Timber, 

a 

2.25 

Nov.   4. 

"  Cash, 

60.00 

Dec.  29. 

"  Draft, 

45.00 

$338.50 

$  34.40 


Bal.  due  B.,  S.,  $  R., 
Jan.  5.     Settled  by  due-bill, 

Baldwin,  Sherwood,  &  Rice 


at 


(5.) 
Mr.  Benjamin  Treat, 

37  chests  Green  Tea, 
41      «      Black  do. 
40      "      Imperial  Tea, 
13  crates  Liverpool  Ware, 


Received  payment, 
N   8 


St.  Louis,  May  1,  1856. 
Bought  of  John  True, 


$25.50 
16.17 
97.75 

169.37 


John  True. 


86  UNITED   STATES  MONEY. 

(6.)  Portland,  May  16,  1857. 

Mr.  J.  C.  Porter, 

Bought  of  Willard  &  Hale, 


17  bbl.  Canada  Flour, 

a*      $  8.25 

50  lbs.  Duponfs  Eagle 
140  "     Sheet  Zinc, 

\  Gunpowder, 

.50 

.081 

120  "    Prussian  Blue, 

■ 

"            .63 

* 

Received 

payment, 

Willard  &  Hale. 

(7.)                                           New  York,  July  11,  1856. 
Mr.  John  Cummings, 

Bought  o/*Lord  &  Secomb, 
97  bbl  Genesee  Flour,          at         $  6.25 

167    "    Philadelphia 
87   "    Baltimore 

do. 

a 

a 

5.95 
6.07 

196   "    Richmond 

do. 

(f 

5.75 

275   "    Howard  St. 

do. 

a 

7.25 

69  bu.  Rye, 
136  "     Virginia  Corn, 

a 
a 

1.16 

.67 

68  "    North  River 

do. 

a 

.76 

169  "     Wheat, 

a 

1.37 

76  tons  Lehigh  Coal, 
89    "    Iron, 

a 
a 

9.67 
69.70 

49  Grindstones, 

u 

3.47 

39  Pitchforks, 
197  ifa£es, 

a 

1.61 
.17 

86  .floes, 

a 

.69 

78  Shovels, 

r< 

1.17 

187  Spades, 
91  Ploughs, 
83  Harrows, 

a 

a 

.85 
11.61 
17.15 

47  Handsaws, 

IK 

3.16 

35  Mill-saws, 

a 

18.15 

47  ctttf.  A$fee£, 

a 

9.47 

57     "    Zeatf, 

E.  T. 

Lowe, 

6.83 

^  17  315  3° 

Received  payment, 

far  Lord  &  Secomb. 

LEDGER   ACCOUNTS.  87 


LEDGER  ACCOUNTS. 

123.  The  principal  book  of  accounts  among  merchants  is 
called  a  ledger.  In  it  are  brought  together  scattered  items 
of  accounts,  often  making  long  columns.  As  a  rapid  way  of 
finding  the  amounts,  accountants  generally  add  more  than  one 
column  at  a  single  operation  (Art.  48).  The  examples  below 
may  be  added  both  by  the  usual  method  and  by  that  which  is 
more  rapid. 


1. 

2. 

3. 

4. 

$   cts. 

$  cts. 

I   cts. 

8   cts. 

3.3  3 

9  1.16 

1  3  4.6  2 

1  2  0.0  3 

9.10 

4  0.4  4 

5  1  1.4  2 

6  3.17 

16.3  7 

6  0.6  6 

2  7.3  4 

1  0  0.0  0 

41.3  0 

9  1.3  1 

3  3.3  5 

7  8  1.1  4 

3  1.7  0 

4.0  0 

6  0.4  0 

1  5  1.6  1 

5  7.71 

6.3  4 

1  4  9.3  1 

3  1.2  1 

3.6  0 

7.19 

9  2  1.3  1 

1  0  5.5  2 

5.0  0 

2  8.3  7 

3  9.16 

3  0  6.0  6 

.5  0 

81.0  0 

1 1  0.0  5 

17.7  5 

.13 

6.0  0 

1  0  4.0  3 

8  5.2  5 

1.5  5 

15.0  0 

3  6  3.4  0 

13.0  0 

7.9  1 

4  3.0  5 

9.0  0 

10.0  0 

1.19 

10.0  3 

11.12 

1  3  5.7  5 

9.0  0 

9  5.0  5 

0.10 

1  1  1.1  5 

13.4  0 

61.18 

17.9  1 

1  6  3.0  5 

3  7.6  5 

1  1  2.1  5 

11.13 

7  1  2.1  5 

4  5.7  5 

1  0  0.6  1 

4  5.9  2 

3  1.15 

5.2  5 

1  6  0.0  0 

17.7  5 

9  5.2  5 

7.7  5 

3  9.4  1 

13.8  5 

2  5.17 

5.3  1 

2  2.3  9 

1  1.0  1 

7  7.2  5 

8.13 

9  2.7  7 

6  2.19 

1  2  3.1  0 

4.2  7 

3  3.0  1 

9  5.17 

1  1  6.5  1 

2.6  4 

6  4.3  2 

1  9.1  0 

2  2  3.0  6 

11.12 

5  9.11 

1  1  1.0  0 

9  6  6.7  5 

41.2  1 

17.8  9 

1  9  5.2  5 

1  0  2.1  5 

3  3.7  9 

7  6  1.1  5 

6  1  3.2  0 

4  1  0  6.3  1 

4  7.3  1 

8  2  3.5  4 

1  0  8.0  9 

6  1  0  2.1  9 

10.0  0 

2  3  8.4  7 

3  3  4.0  5 

7  1  1  9.8  1 

88  KEDUCTION   OF   COMPOUND  NUMBERS. 


COMPOUND  NUMBERS. 

124.  A  Compound  Number  is  a  collection  of  concrete 
units  of  different  denominations  ;  as,  5  pounds  and  6  ounces ;  4 
feet  and  5  inches. 

125.  A  scale  expresses  the  law  of  relation  between  the 
different  units  of  a  number. 

The  different  units  of  simple  numbers  have  a  uniform  tenfold 
increase  from  lower  to  higher  orders,  and  a  like  decrease  from 
higher  to  lower  orders.  They,  therefore,  are  said  to  have  a 
uniform  scale. 

In  compound  numbers,  the  names  of  different  measuring 
units  (Art.  9)  are  included  in  the  expression  of  a  single 
quantity,  so  that  the  relation  of  the  units  of  one  order  to  those 
of  another  is  that  of  a  varying  scale  ;  as  in  the  expression  of 
pounds,  shillings,  and  pence,  it  is  4,  12,  and  20. 

REDUCTION  OF  COMPOUND  NUMBERS. 

126.  Reduction  is  the  process  of  changing  numbers  from 
one  denomination  to  another,  without  altering  their  values. 

It  is  of  two  kinds,  Reduction  Descending  and  Reduction  As- 
cending. 

Reduction  Descending  is  changing  numbers  of  a  higher  de- 
nomination to  a  lower  denomination ;  as  pounds  to  shillings,  &c. 
It  is  performed  by  multiplication. 

Reduction  Ascending  is  changing  numbers  of  a  lower  denom- 
ination to  a  higher  denomination  j  as  farthings  to  pence,  &c. 
It  is  the  reverse  of  Reduction  Descending,  and  is  performed  by 
division. 

ENGLISH  MONEY. 

127.  English  or  Sterling  Money  is  the  currency  of  Eng- 
land. 

Table. 


4  Farthings  (qr.  or  far.)         make         1  Penny,  d. 

12  Pence  "  1  Shilling,  s. 

20  Shillings  "  1  Pound,  £. 


REDUCTION  OP  COMPOUND  NUMBERS.  89 


far. 

d. 

4 

= 

1 

s. 

48 

= 

12 

= 

1 

£. 

960 

= 

240 

= 

20 

= 

1 

Note  1.  —  The  symbol  £.  stands  for  the  Latin  word  libra,  signifying  a 
pound;  s.  for  solidus,  a  shilling;  d.  for  denarius,  a  penny;  qr.  for  quadrans, 
a  quarter. 

Note  2.  —  Farthings  are  sometimes  expressed  in  a  fraction  of  a  penny; 
thus,  1  far.  =  i  d.;  2  far.  =  £  d.;  3  far.  =  |  d. 

Note  3.  —  The  term  sterling  is  probably  from  Easterling,  the  popular  name 
of  certain  early  German  traders  in  England,  whose  money  was  noted  for  the 
purity  of  its  quality. 

Note  4.  —  The  English  coins  consist  of  the  five-sovereign  piece,  the  double- 
sovereign,  the  sovereign,  and  the  half-sovereign,  made  of  gold;  the  crown,  the 
half-crown,  the  shilling,  the  six-pence,  the  four-pence,  the  three-pence,  the  two- 
pence, the  one-and-a-half-pence,  and  the  penny,  made  of  silver ;  the  penny,  the 
half-penny,  the  farthing,  and  the  half-farthing,  made  of  copper. 

The  sovereign  represents  the  pound  sterling,  whose  legal  value  in  United 
States  money  is  $4.84.  The  value  of  the  English  guinea  is  21  shillings  ster- 
ling. The  guinea,  the  five-guinea,  the  half-guinea,  the  quarter-guinea,  and  the 
seven-shilling  piece,  are  no  longer  coined. 

The  guinea  is  so  called  because  the  gold  of  which  the  first  guineas  were 
made  was  brought  from  Guinea,  in  Africa. 

The  English  gold  coins  are  now  made  of  11  parts  of  pure  gold,  and  1  part  of 
copper,  or  some  other  alloy ;  and  the  silver  coin,  of  37  parts  of  pure  silver,  and 
3  parts  of  copper. 

The  present  standard  weight  of  the  sovereign  is  223i|l  grains  Troy;  the 
crown,  440^f  grains ;  the  copper  penny,  291|  grains. 

128.  To  change  numbers  expressed  in  one  or  more  de- 
nominations to  their  equivalents  in  one  or  more  other  denom- 
inations. 

Ex.  1.     In  48£.  12s.  7d.  2far.  how  many  farthings? 

operation.  "VVe  multiply  the  48  by  20,  be- 

4  8  £.  1  2  s.  7  d.  2  far.  cause  20  shillings  make  1  pound, 
and  to  this  product  we  add  the  12 
shillings  in  the  question,  and  obtain 
shillings.  972  shillings.     We  then  multiply 

by  12,  because  12  pence  make  1 
shilling,  and  to  the  product  we  add 
P  the    7    pence,  and    obtain   11671 

pence.     Again,  we  multiply  by  4, 
Ans.  4  6  6  8  6  farthings.  because  4  farthings  make  1  penny, 

and  to  this  product  we  add  the  2 
farthings,  and  obtain  46686  farthings,  the  answer  sought. 
8* 


20 

972 
12 

11671 

4 

90  REDUCTION   OP  COMPOUND  NUMBERS. 

Ex.  2.     In  46686  farthings  how  many  pounds  ? 

operation.  We  divide  by  4,  because  4  far- 

4  )  4  6  6  8  6  far.  things  make  1  penny,  and  the  re- 

i  o  \  -i  t  nr,  t  i    o  e  sm1t  is  11671  pence,  and  2  farthings 

1  I  )Hb71  d.  Z  tar.  remaining.     We  then  divide  by  12 

20)  972  s.  7  d.  because  12  pence  make  1  shilling, 

and  the  result  is  972  shillings,  and 

4  8  £.  1  2  s.  7  pence  remaining.     Lastly,  we  di- 

Ans.  48£.  12s.  7d.  2  far.       T»  hr  .20>  because   20  shillings 

make  1  pound,  and  the  result  is 
48  pounds,  and  12  shillings  remaining.  By  annexing  to  the  last 
quotient  the  several  remainders,  we  obtain  48£.  12s.  7d.  2far.  as  the 
required  result. 

From  these  illustrations,  for  the  two  kinds  of  reduction,  we  deduce 
the  following 

Rule.  —  For  Reduction  Descending.  Multiply  the  highest 
denomination  given  by  the  number  of  units  required  of  the  next  lower 
denomination  to  make  one  in  the  denomination  multiplied.  To  this 
product  add  the  corresponding  denomination  of  the  multiplicand,  if 
there  be  any.  Proceed  in  this  way,  till  the  reduction  is  brought  to  the 
denomination  required. 

For  Reduction  Ascending.  Divide  the  lowest  denomination 
given  by  the  number  of  units  required  of  that  denomination  to  make  one 
of  the  next  higher.  The  quotient  thus  obtained  divide  as  beforehand 
so  proceed  until  it  is  brought  to  the  denomination  required.  The  last 
quotient,  with  the  several  remainders,  if  there  be  any,  annexed,  will  be 
the  answer. 

Examples. 

3.  In  127£.  15s.  8d.  how  many  farthings  ? 

4.  In  122672  farthings  how  many  pounds  ? 

5.  How  many  farthings  in  28£.  19s.  lid.  3  far.  ? 

6.  How  many  pounds  in  27839  farthings  ? 

7.  In  37  8£.  how  many  pence  ? 

8.  In  90720  pence  how  many  pounds  ? 

9.  Reduce  967  guineas  to  pounds. 
10.  Reduce  1015£.  7s.  to  guineas. 


AVOIRDUPOIS  WEIGHT. 

129.  Avoirdupois  or  Commercial  Weight  is  used  in  weigh- 
ing almost  every  kind  of  goods,  and  all  metals  except  gold  and 
silver. 


REDUCTION   OP   COMPOUND   NUMBERS.  91 


Table. 


16  Drams  (dr.) 

16  Ounces 

25  Pounds 

4  Quarters 

20  HundredWeight 

4r. 

oz. 

16       = 

1 

256       = 

16 

make  1  Ounce,  oz. 

"  1  Pound,  lb. 

"  1  Quarter,  qr. 

"  1  Hundred  Weight,  cwt. 

M  1  Ton,  T. 


lb. 

1        qr. 

6400   =     400   =     25  =    1      cwt. 

25600   ==    1600   =    100  =    4   =    1      T. 

51200   =   32000   =   2000  =   80   =   20   =*  1 


Note  1.  —  The  oz.  stands  for  onza,  the  Spanish  for  ounce,  and  in  cwt.  the  c 
stands  for  centum,  the  Latin  for  one  hundred,  and  wt  for  weight. 

Note  2.  —  The  laws  of  most  of  the  States,  and  common  practice  at  the 
present  time,  make  25  pounds  a  quarter,  as  given  in  the  table.  But  formerly, 
28  pounds  were  allowed  to  make  a  quarter,  112  pounds  a  hundred,  and  2240 
pounds  a  ton,  as  is  still  the  standard  of  the  United  States  government  in  col- 
lecting duties  at  the  custom-houses. 

Note  3.  —  The  term  avoirdupois  is  from  the  French  avoir  dupoid,  signifying 
to  have  weight. 

Note  4.  —  The  standard  avoirdupois  pound  of  the  United  States  is  the 
weight,  taken  in  the  air,  of  27I70°01050  cubic  inches  of  distilled  water,  at  its  max- 
imum density,  or  when  at  a  temperature  of  39^  degrees  Fahrenheit,  the  ba- 
rometer being  at  30  inches.  It  is  the  same  as  the  Imperial  pound  avoirdupois 
of  Great  Britain,  which  is  the  weight  of  27^§j  cubic  inches  of  distilled  water 
at  the  temperature  of  62  degrees. 


Examples. 

1.  In  165T.  13cwt.  3qr.  191b.  14oz.  how  many  ounces  ? 

2.  In  5302318  ounces  how  many  tons  ? 

3.  If  a  load  of  hay  weigh  3T.  16cwt.  2qr.  181b.,  required 
the  weight  in  ounces. 

4.  In  122688  ounces  how  many  tons  ? 

5.  Required  the  number  of  drams  in  2T.  17cwt.  3qr.  161b. 
15oz.  13dr. 

6.  In  1482749  drams  how  many  tons  ? 

7.  What  is  the  value  of  7T.  17cwt.  at  7  cents  per  pound? 

Ans.  $  1099.00. 

8.  What  will  19cwt.  3qr.  201b.  of  sugar  cost  at  9  cents  per 
pound?  Ans.  $179.55. 


92  REDUCTION   OP   COMPOUND   NUMBERS. 

TROY  OR  MINT  WEIGHT. 

130.  Troy  or  Mint  Weight  is  the  weight  used  in  weighing 
gold,  silver,  jewels,  and  liquors ;  and  in  philosophical  experi- 
ments. 

Table. 


24  Grains  (gr.) 
20  Pennyweights 
12  Ounces 

make 

a 
U 

1  Pennyweight, 
1  Ounce, 
1  Pound, 

pwt. 
oz. 

lb. 

gr. 

24              — 

pwt. 

1 

oz. 

480             — 

20 

=         1 

lb. 

5760             = 

240 

=           12         = 

1 

Note  1.  —  Troy  weight  was  introduced  into  Europe  from  Cairo  in  Egypt, 
in  the  12th  century,  and  was  first  adopted  in  Troyes,  a  city  in  France,  where 
great  fairs  were  held,  whence  it  may  have  had  its  name. 

Note  2.  —  A  grain  or  corn  of  .wheat,  gathered  out  of  the  middle  of  the  ear, 
was  the  origin  of  all  the  weights  used  in  England.  Of  these  grains,  32,  well 
dried,  were  to  make  one  pennyweight.  But  in  later  times  it  was  thought 
sufficient  to  divide  the  same  pennyweight  into  24  equal  parts,  still  called 
grains,  being  the  least  weight  now  in  use,  from  which  the  rest  are  computed. 

Note  3.  —  Diamonds  and  other  precious  stones  are  weighed  by  what  is 
called  Diamond  Weight,  of  which  16  parts  make  1  grain ;  4  grains,  1  carat. 
1  grain  Diamond  Weight  is  equal  to  f  grains  Troy,  and  1  carat  to  3}  grains 
Troy.  In  weighing  pearls,  the  pennyweight  is  divided  into  30  grains  instead 
of  24,  so  that  1  pearl  grain  is  equal  to  |  grains  Troy.  The  carat  as  a  weight 
must  not  be  confounded  with  the  assay  carat,  a  term  whose  use  is  to  indicate 
a  proportional  part  of  a  weight,  as  in  expressing  the  fineness  of  gold,  each 
carat  means  a  twenty-fourth  part  of  the  entire  mass  used.  Thus,  pure  gold  is 
termed  24  carat  gold,  and  gold  that  is  not  pure  is  termed  18  carat  gold,  20 
carat  gold,  &c,  as  its  mass  may  be  18  twenty-fourths,  20  twenty-fourths,  &c. 
pure  gold.  Each  assay  carat  is  subdivided  into  4  assay  grains,  and  each 
assay  grain  into  4  assay  quarters. 

Note  4.  —  The  Troy  pound,  the  standard  unit  of  weight  adopted  by  the 
United  States  Mint,  is  the  same  as  the  Imperial  Troy  pound  of  Great  Britain, 
and  is  equal  to  the  weight,  taken  in  the  air,  of  22^^  cubic  inches  of  distilled 
water,  at  its  maximum  density,  the  barometer  being  at  30  inches. 

Examples. 

1.  How  many  grains  in  281b.  lloz.  12pwt.  15gr.  Troy? 

2.  In  166863  grains  Troy  how  many  pounds  ? 

3.  If  a  silver  pitcher  weigh  31b.  10oz.,  what  is  its  weight 
in  grains  ? 

4.  How  many  pounds  Troy  in  22080  grains  ? 


REDUCTION   OF   COMPOUND   NUMBERS.  93 

5.  What  is  the  value  of  731b.  lloz.  of  standard  silver  at 
$  0.062  per  pennyweight  ? 

6.  How  many  pounds  of  standard  silver  can  be  purchased 
for  $  1099.88,  at  the  rate  of  $0,062  per  pennyweight? 

7.  A  Californian  has  571b.  7oz.  of  pure  gold.     What  is  its 
value  at  $  20.593±  per  ounce  ?  Ans.  $  14229.901f 

8.  What  is  the  value  of  a  mass  of  standard  gold  weighing 
191b.  6oz.  16pwt.  at  93  cents  per  pennyweight? 

Ans.  $4367.28. 

9.  I  have  a  lump  of  pure  silver  weighing  131b.  9oz.     What 
is  its  value  at  $  1.385  &  per  ounce  ?  Ans.  $  228.640  J. 

APOTHECARIES'  WEIGHT. 

131  •     Apothecaries'  Weight  is  used  in  mixing  medical  pre- 
scriptions. 

Table. 


20  Grains  (gr 

0 

make 

1  Scruple, 

sc.  or  £). 

3  Scruples 

u 

1  Dram, 

dr.  or  5  • 

8  Drams 

ii 

1  Ounce, 

oz.  or  §  . 

12  Ounces 

a 

1  Pound, 

lb.  or  ft,. 

gr. 

sc. 

20           = 

1 

dr. 

60           = 

3 

=s 

1 

oz. 

480  =  24  =  8  =  1  lb. 

5760  =  288  =  96  =  12  =  1 

Note  1.  —  In  this  weight  the  pound,  ounce,  and  grain  are  the  same  as  in 
Troy  Weight. 

Note  2.  —  Medicines  are  usually  bought  and  sold  by  Avoirdupois  Weight. 

Note  3.  —  In  estimating  the  weight  of  fluids,  45  drops,  or  a  common  tea- 
spoonful,  make  about  1  fluid  dram ;  2  common  table-spoonfuls,  about  1  fluid 
ounce;  a  wineglassful,  about  1^  fluid  ounces;  and  a  common  teacupful,  about 
4  fluid  ounces. 

Examples. 

1.  In  23ft)  9 §  05  29  13gr.  how  many  grains? 

2.  In  136853  grains  how  many  pounds  ? 

3.  How  many  scruples  in  231b.  ? 

4.  How  many  pounds  in  6624  scruples? 

5.  In  47tb  OS  05  19  19gr.  how  many  grains? 

6.  In  270759  grains  how  many  pounds  ? 


94  REDUCTION   OF   COMPOUND   NUMBERS. 

7.  A  physician  bought  1  pound  of  ipecacuanha  for  $  1.80, 
and  retailed  it  out  in  doses  of  5  grains,  at  12£  cents  each.  How 
much  did  he  get  for  it  over  the  cost  ?  Ans.  $  142.20. 

Avoirdupois,  Troy,  and   Apothecaries'  Weight    Com- 
pared. 

132.  The  relative  value  of  the  pound,  and  its  subdivisions, 
of  the  several  weights,  in  Troy  grains,  and  in  denominations  of 
each  other,  is  shown  in  the  following 

Table. 

1  lb.  Av.  =  7000  gr.  Tr.  =  lib.  2oz.  llpwt.  16gr.  Tr. 

1  lb.  Tr.  or  Ap.  —  57G0  "  —  13oz.  2{i|dr.  Av. 

1  oz.  Tr.  or  Ap.  =  480  "  —  loz.  lftdr.  Av. 

1  oz.  Av.  =  43  7 J  "  =  18pwt.  5jgr.  Tr. 

1  dr.  Ap.  =  60  "  =  2T3745dr.  Av. 

1  dr.  Av.  =  27J£  "  —  lpwt.  SUgr.  Tr. 

1  pwt.  Tr.  =  24  "  =  Iffdr.  Av. 

1  sc.  Ap.  =  20  "  =  }f|dr.  Av. 

1  gr.  Tr.  or  Ap.  =  1  "  =  g3f5dr.  Av. 

Note.  —  To  change  a  quantity  from  one  weight  to  its  equivalent  in  another 
weight,  reduce  the  yivcn  quantity  to  Troy  grains,  and  Hienjind  their  value  in  de- 
nominations of  die  weight  required. 

Examples. 

1.  Change  131b.  6oz.  Avoirdupois  weight  to  Troy  weight. 

2.  Change  161b.  3oz.  lpwt.  lgr.  Troy  weight  to  Avoirdupois 
weight. 

3.  Change  31b.  8oz.  lOpwt.  to  drams  of  Apothecaries'  weight. 

Ans.  356dr. 

4.  Change  356  drams  Apothecaries'  weight  to  Troy  weight. 

Ans.  31b.  8oz.  lOpwt. 

5.  An  apothecary  bought  by  Avoirdupois  weight  21b.  8oz.  of 
quinine  at  $  2.40  per  ounce,  which  he  retailed  at  20  cents  a 
scruple.     What  was  his  gain  on  the  whole  ? 

6.  If  I  should  buy  by  Avoirdupois  weight  121b.  of  opium  at 
37^  cents  per  ounce,  and  sell  it  by  Troy  weight  at  40  cents  per 
ounce,  should  I  gain  or  lose  by  so  doing  ?         Ans.  Lose  $  2. 


REDUCTION   OF   COMPOUND   NUMBERS. 


95 


LINEAR   OR  LONG  MEASURE. 

133.     Linear  or  Long  Measure  is  used  in  measuring  dis- 
tances in  any  direction. 

Table. 


12    Inches  (in.)                make     1 

Foot,                                         ft. 

3    Feet 

it 

1 

Yard,                                        yd. 

5J  Yards,  or  16$  Feet, 

u 

1 

Rod,  or  Pole,                          rd. 

40    Rods 

u 

1 

Furlong,                                   fur. 

8    Furlongs,  or  320  Rods, 

a 

1 

Mile,                                        m. 

3    Miles 

a 

1 

League,                                  lea. 

69J  Miles  (nearly) 

a 

1 

Degree  on  the  equator,  deg.  or  °. 

360    Degrees 

a 

1 

Great  Circle  of  the  Earth. 

in.                           ft. 
12       =              1 

yd. 

36         =                3         mm 

1 

rd. 

198       =           16|-     = 

&h 

=             1                  fur. 

7920       =         660       = 

220 

=          40       =        1                  m. 

63360       =       5280       = 

1760 

=       320       =       8       =       1 

Note  1.  — 12  lines  make  1  inch;  4  inches,  1  hand;  6  feet,  1  fathom;  120 
fathoms,  1  cable-length;  7  J  cable-lengths,  1  mile;  i  of  a  degree  of  the  circum- 
ference of  the  earth,  1  knot,  or  geographical  mile,  equal  to  l|i  statute  miles. 

Note  2.  —  The  yard  adopted  by  the  United  States  government  as  the 
standard  unit  of  linear  measure  is  the  same  as  the  imperial  yard  of  Great 
Britain,  which,  as  compared  with  a  pendulum  vibrating  seconds  in  the  latitude 
of  London,  the  pendulum  moving  in  a  vacuum,  at  the  level  of  the  sea,  and  at 
the  temperature  of  62°  Fahrenheit,  should  bear  the  proportion  of  36  to  39^3 
inches.  A  metre,  the  unit  of  linear  measure,  as  established  by  the  French 
government,  is  equal  to  about  39^  English  inches. 

Note  3.  —  The  English  statute  mile  is  the  same  as  that  of  the  United 
States,  but  that  of  other  countries  differs  in  value  from  it;  as  the  German 
short  mile  is  equal  to  6857  yards,  or  about  3j90  English  miles;  the  German  long 
mile,  to  10125  yards,  or  about  5|  English  miles;  the  Prussian  mile,  to  8237 
yards,  or  about  4jTg  English  miles;  the  Spanish  common  league,  to  7416  yards, 
or  about  4|  English  miles;  the  Spanish  judicial  league,  to  4635  yards,  or  about 
2|  English  miles. 

.  Note  4.  —  A  degree  of  longitude  is  ^  of  any  circle  of  latitude.  As  the 
circles  of  latitude  diminish  in  length,  the  degrees  of  longitude  vary  in  length 
under  different  parallels  of  latitude.  Thus,  under  the  equator,  the  length  of  a 
degree  of  longitude  is  about  69§  statute  miles;  at  25°  of  latitude  62T7S  miles;  at 
40°  of  latitude,  53  miles;  at  42°  of  latitude,  5l£  miles;  at  49°  of  latitude,  45£ 
miles;  at  60°,  34/§  miles. 


96  REDUCTION   OF  COMPOUND  NUMBERS. 

Examples. 

1.  In  96deg.  56m.  7fur.  32rd.  12ft.  6in.  how  many  inches  ? 

2.  In  424320486  inches  how  many  degrees  ? 

3.  How  many  feet  in  79  miles  ? 

4.  Required  the  miles  in  417120  feet. 

5.  How  many  inches  in  396  furlongs  ? 

6.  Required  the  furlongs  in  3136320  inches  ? 

7.  How  many  inches  from  Haverhill  to  Boston,  the  distance 
being  30  miles  ? 

8.  Required  the  miles  in  1900800  inches. 

CLOTH  MEASURE. 

134.     Cloth  Measure  is  used  in  measuring  cloth,  ribbons, 
lace,  and  other  articles  sold  by  the  yard  or  ell. 

Table. 


24  Inches  (in 

) 

make 

1  Nail, 

na. 

4    Nails 

K 

1  Quarter  of  a 

yard, 

qr. 

4    Quarters 

u 

1  Yard, 

yd. 

3    Quarters 

it 

1  Ell  Flemish, 

E.  F. 

5    Quarters 

U 

1  EU  English, 

E.  E. 

in. 

na. 

2i       = 

1 

qr. 

9         — 

4 

— 

1 

E.  F. 

27         = 

12 

= 

3 

=        1 

yd. 

36         = 

16 

= 

4 

=       H     = 

1 

E.E. 

45         = 

20 

= 

5 

=       i|     = 

li 

=        1 

Note  1.  —  The  Ell  French  is  6  quarters;  the  Ell  Scotch,  4qr.  l|in. 
Note  2.  —  Cloth  measure  is  a  species  of  linear  measure,  and  the  yard  and 
inch  are  the  same  in  both. 

Examples. 

1.  In  17yd.  3qr.  2na.  how  many  nails  ? 

2.  In  286  nails  how  many  yards  ? 

3.  In  365yd.  lqr.  3na.  how  many  nails  ? 

4.  In  5847  nails  how  many  yards  ? 

5.  In  71E.  E.  4qr.  how  many  nails  ? 

6.  In  1436  nails  how  many  ells  English  ? 

7.  What  cost  47yd.  3qr.  of  silk  velvet  at  $  1.25  per  quarter  ? 

8.  A  merchant  bought  a  roll  of  cloth  containing  31£E.  E. 
and  paid  for  it  at  the  rate  of  $  3  per  yard.  What  did  it  cost 
him?  Ans.  $117. 


REDUCTION   OP   COMPOUND   NUMBERS. 


97 


135. 

kinds. 


SUEFACE   OR  SQUARE  MEASURE. 
Square  Measure  is  used  in  measuring  surfaces  of  all 


Table. 


Square  inches  (sq.  in.)  make 
Square  feet  " 


144 
9 

30£  Square  yards 
40    Square  rods 

4    Roods 
640    Acres 

in. 

144  = 

1596  = 

39204  = 

1568160  = 

6272640  = 


Square  foot, 

Square  yard, 

Square  rod  or  pole, 

Rood, 

Acre, 

Square  mile, 


ft. 

yd. 

P- 
R. 

A. 

S.M. 


ft. 
1 

9  = 
272i  = 

10890  = 
43560  = 


yd- 
1 
30£  = 
1210  = 

4840  = 


P- 
1 

40  = 

160  = 


R. 
1 

4 


4014489600  =  27878400  =  3097600  =  102400  =  2560 


A. 

1    S.M. 
640  =  1 


Square 
foot. 

Note.  —  A  square  is  a  figure  bounded  by  four  equal  lines,  perpendicular  to 
each  other. 

When  the  four  lines  are  each  1  foot  in  length,  the  space  enclosed  is  1  square 
foot;  when  1  yard  in  length,  1  square  yard;  when  1  rod  in  length,  1  square 
rod;  and  so  for  any  other  dimension. 

3ft.  =  1yd. 

In  this  diagram  the  large  square  repre- 
sents a  square  yard,  and  each  of  the  smaller 
squares  within  it  represents  one  square  foot. 
Now,  since  therfc  are  three  rows  of  small 
squares,  and  three  square  feet  in  each  row, 
there  will  be  3  times  3  =  9  sq.  ft.  in  the 
large  square.  But  the  large  square  is  3ft. 
in  length  and  3ft.  in  breadth ;  hence, 

To  find  the  contents  of  a  square,  multiply  its  length  by  its  breadth. 

Examples. 

1.  In  57A.  3R.  27p.   21yd.  8ft.  57in.    how  many  square 
inches  ? 

2.  In  363331893  square  inches  how  many  acres? 
How  many  square  feet  in  25  acres  ? 
How  many  acres  in  1089000  square  feet? 
How  many  square  rods  in  365  square  miles  ? 
How  many  square  miles  in  37376000  square  rods  ? 
How  many  acres  in  12345678  square  inches? 

Ans.  1A.  3R.  34p.  27yd.  4ft.  54in. 
N    9 


3, 

4. 

5. 
0. 

7. 


98  REDUCTION   OF   COMPOUND   NUMBERS. 

8.  Bought  39 A.  2R.  16p.  of  land  for  $3.75  per  square  rod, 
and  sold  the  same  for  $0.25  per  square  foot.  What  did  I 
gain  by  my  bargain  ?  Ans.  $  407,484.00. 

SURVEYORS'   MEASURE. 

136.  This  measure  is  used  by  surveyors  in  measuring 
land,  roads,  &c. 


Table. 


7jfc  Inches  (in.) 

make 

1  Link 

25      Links 

u 

1  Pole, 

00      Links,  4  Poles,  or  66  Feet 

U 

1  Chain, 

10      Chains 

u 

1  Furlong, 

8      Furlongs,  or  80  Chains, 

u 

1  Mile, 

Inches.                       Link. 

7  92      _                         l 

Tole. 

198        =             25        = 

1 

Chain. 

792        =           100        = 

4       = 

1                 Furlong. 

7920        =        1000        = 

40       = 

10       =       1 

1. 

p- 

ch. 
fur. 
m. 


Mile. 
63360       =        8000        =       320       —       80       =       8       ==     1 

Note  1.  —  Guntcr's  chain,  in  length  4  poles,  or  66  feet,  and  divided  into  100 
links,  is  that  mostly  used  in  ordinary  land  surveys ;  but  in  locating  roads,  and 
like  public  works,  an  engineer's  chain  is  usually  100  feet  in  length,  containing 
120  links,  each  10  inches  long. 

Note  2.  —  A  section  of  government  lands  is  1  square  mile,  or  640  acres. 
An  acre,  as  a  square  piece  of  land,  will  measure  on  each  side  about  209  feet 
or  70  paces.  625  square  links  make  1  square  rod  or  pole ;  16  square  rods  make 
1  square  chain,  and  10  square  chains  make  1  acre. 

Note  3.  —  A  rod  or  pole  is  sometimes  called  a  perch,  and  each  of  the  names 
given  to  this  measure  is  expressive  of  the  instrument  by  which  it  was  formerly 
measured. 

Examples. 

1.  How  many  links  in  46m.  3fur.  och.  251.  ? 

2.  In  371525  links  how  many  miles  ? 

3.  In  97m.  Ofur.  how  many  links  ? 

4.  In  776000  links  how  many  miles  ? 

5.  The  extent  of  a  certain  farm  is  found,  by  survey,  to  be 
1377  square  chains  (Note  2).  How  many  acres  does  it  con- 
tain ?  Ans.  137A.  2E.  32p. 

6.  What  will  be  the  cost  of  a  field  measuring  2,126,250 
square  links,  at  $  80  per  acre  ?  Ans.  $  1701.00. 


REDUCTION   OF   COMPOUND   NUMBERS. 


99 


CUBIC  OR  SOLID  MEASURE. 

137.  Cubic  or  Solid  Measure  is  used  in  measuring  such 
bodies  or  things  as  have  length,  breadth,  and  thickness;  as 
timber,  stone,  &c. 

Table. 


1728  Cubic  inches  (cu.  in.) 

make  1  Cubic  foot, 

cu.  ft. 

27       " 

feet 

u 

1       "      yard, 

cu.  yd. 

40       " 

feet 

(4 

1  Ton, 

T. 

16       " 

feet 

ft 

1  Cord  foot, 

eft. 

8  Cord  feet,  or ") 
128  Cubic  feet,     } 

a 

1  Cord  of  wood, 

C. 

in. 
1728 

— 

ft. 
1 

yd. 

46656 

= 

27 

= 

1                             T. 

69120 

= 

40 

= 

m          =           1 

c. 

221184 

= 

128 

= 

4f         =         S\ 

=       1 

Note  1.  —  A  pile  of  wood  8ft.  in  length,  4ft.  in  breadth,  and  4ft.  in  height, 
contains  a  cord. 

Also,  one  ton  of  timber,  as  usually  surveyed,  contains  50j9Q20  cubic  or  solid 
feet. 

Sawed  timber,  joists,  plank,  and  scantlings  are  now  generally  bought  and 
sold  by  what  is  called  board  measure. 

Note  2.  —  A  cube  is  a  solid  bounded  by  six  square  and  equal  sides. 

If  the  cube  is  1  foot  long,  1  foot  wide,  and  1  foot  high,  it  is  called  a  cubic  or 
solid  foot.  If  the  cube  is  S  feet  long,  3  feet  wide,  and  3  feet  thick,  it  is  called 
a  cubic  or  solid  yard. 

Now,  since  each  side  of  a  cubic  yard, 
as  represented  in  the  diagram,  contains 
9  sq.  ft.  of  surface  (Art.  93),  it  is  plain, 
if  a  block  be  cut  off  from  one  side,  one 
foot  thick,  it  can  be  divided  into  9  solid 
blocks,  with  sides  1  foot  in  length, 
breadth,  and  thickness,  and  therefore 
will  contain  9  solid  feet ;  and  since  the 
whole  block  or  cube  is  three  feet  thick, 
it  must  contain  3  times  9  =  27  feet; 
for  3ft.  x  3ft.  X  3ft.  =  27  solid  feet. 
Hence, 

To  find  the  contents  of  a  cubic  or  solid  body,  multiply  its  length, 
breadth,  and  thickness  together. 

Note  3.  —  A  cubic  foot  of  distilled  water  at  the  maximum  density,  at  the 
level  of  the  sea,  and  the  barometer  at  30  inches,  is  equal  in  weight  to  62£lb. 
or  lOOOoz.  avoirdupois. 


^1 

/-  / 

*s  •>  ^  4\ 

•ff-/^    /       „rHBIIIII 

d 

Bll 

llli 

II 

111 

lliiiillii 

CO 

111 

ill  ill 

ipr 

3  ft.  =  1  yd. 


100 


REDUCTION   OF   COMPOUND   NUMBERS. 


Note  4.  —  A  cubic  foot  of  lead  weighs  708|lb. ;  of  brass,  534|lb. ;  of  cop- 
per, 5551b.;  of  wrought-iron,  486flb.;  of  cast-iron.  450£lb.;  of  marble, 
1711b.;  of  granite,  1651b.;  of  clay,  1301b.;  of  common  soil,  1241b.;  of 
bricks,  1241b.;  of  sand,  951b.;  of  sea-water,  64^1^;  of  oak  wood,  551b.; 
of  Anthracite  coal,  541b.;'  of  Bituminous  coal,  501b.;  of  red-pine  wood, 
421b. ;  and  of  white-pine  wood,  301b. 

Examples. 

1.  In  29  cords  of  wood  how  many  solid  inches? 

2.  In  6414336  cubic  inches  how  many  cords? 

3.  In  19  tons  of  timber  how  many  solid  inches  ? 

4.  How  many  tons  of  timber  in  1313280  cubic  inches  ? 

5.  How  many  cubic  feet  of  wood  in  128  cords? 

6.  How  many  cords  of  wood  in  16384  cubic  feet  ? 

7.  How  many  cubic  feet  in  a  pile  of  wood,  40  feet  long,  4 
feet  wide,  and  7  feet  high  ? 

8.  How  many  cords  of  wood  in  8650  cubic  feet  ? 

Ans.  67  cords,  74  cubic  feet. 

9.  How  many  cubic  feet  in  a  granite  block,  17  feet  long,  11 
feet  wide,  and  9  feet  high  ?  Ans.  1683  cubic  feet. 


LIQUID   OR  WINE  MEASURE. 

138.  Liquid  or  Wine  Measure  is  used  in  measuring  all 
kinds  of  liquids,  except,  in  some  places,  beer,  ale,  porter,  and 
milk. 

Table. 


4  Gills  (g 

i) 

2  Pints 

4  Quarts 

63  Gallons 

2  Hogsheads 

2  Pipes 

gi- 

pt. 

4       = 

1 

8      = 

2 

32      = 

8 

2016      = 

504 

4032      = 

1008 

8064      = 

2016 

take 

K 

a 
u 

1  Pint, 
1  Quart, 
1  Gallon, 
1  Hogshead, 

pt. 
qt. 
gal. 
hhd. 

u 
u 

1  Pipe,  or 
1  Tun, 

Butt, 

pi. 
tun. 

qt. 

1 

gal. 

4 

=            1 

hhd. 

252 

=        63     = 

:        1 

pi. 

504 

=       126     = 

:         2        = 

1 

tun. 

1008 

=      252     = 

4     == 

2 

=    1 

EEDUCTION   OF   COMPOUND   NUMBERS.  101 

Note  1.  —  Bylaws  of  Massachusetts,  32  gallons  make  1  barrel.  In  some 
States  3l£  gallons,  and  in  others  from  28  to  32  gallons,  make  1  barrel.  42  gal- 
lons make  1  tierce,  and  2  tierces  make  1  puncheon. 

Note  2.  —  The  term  hogshead  is  often  applied  to  any  large  cask  that  may 
contain  from  50  to  120  gallons,  or  more. 

Note  3.  —  The  Standard  Unit  of  Liquid  Measure  adopted  by  the  govern- 
ment of  the  United  States  is  the  Winchester  Wine  Gallon,  which  contains  231 
cubic  inches,  and  is  of  a  capacity  to  hold  8f^jlb.  Avoirdupois  of  distilled 
water,  at  its  maximum  density,  weighed  in  air,  the  barometer  being  at  30 
inches.  It  has  the  name  Winchester,  from  its  standard  having  been  formerly 
kept  at  Winchester,  England.  The  Imperial  Gallon,  now  adopted  in  Great 
Britain,  contains  277^^  cubic  inches;  so  that  6  Winchester  gallons  make 
about  5  Imperial  gallons. 

Note  4.  —  1  gallon  of  alcohol  weighs  71b. ;  of  camphene,  7|lb. ;  of  proof 
spirits,  7]Jlb. ;  of  spirits  of  turpentine,  7jglb. ;  of  sperm  oil,  7|lb. ;  of  olive 
oil,  7£lb. ;  of  linseed  oil,  7f lb. ;  and  of  molasses,  llflb. 

Note  5.  —  The  fluid  measure  of  apothecaries,  used  by  them  in  measuring 
liquids  of  medical  prescriptions,  divides  the  gallon  (marked  Cong.)  into  8 
pints  (0.);  the  pints  into  16  fluid  ounces  (f^  );  the  fluid  ounces  into  8  fluid 
drams  (f5 );  and  the  fluid  drams  into  60  minims  (m)  or  drops.  The  abbrevi- 
ation Cong,  stands  for  congiarium,  the  Latin  for  gallon,  and  the  O.  is  the  initial 
of  octans,  the  Latin  for  an  eighth,  the  pint  being  an  eighth  of  a  gallon. 

Examples. 

1.  In  57T.  3hhd.  50gal.  3qt.  how  many  pints  ? 

2.  In  116830  pints  how  many  tuns? 

3.  Reduce  96hhd.  47gal.  2qt.  to  gills. 

4.  How  many  hogsheads  in  195056  gills  ? 

5.  What  cost  40  hogsheads  of  wine  at  $  0.37^  per  pint  ? 

6.  How  much  may  be  gained  by  buying  2  hogsheads  of 
molasses,  at  40  cents  a  gallon,  and  selling  it  at  12  cents  a 
quart?  Ans.  $10.08. 

BEER  MEASURE. 

139.  Beer  Measure  is  used  in  measuring  beer,  ale,  porter, 
and  milk. 

Table. 


2  Pints  (pt.) 
4  Quarts 
54  Gallons 

make 

(4 

u 

1  Quart, 
1  Gallon, 
1  Hogshead, 

qt. 

gal. 

hhd. 

pt. 
2 
8 
432 

qt. 

=                   1 

=               4 

=          216 

9* 

gal. 
=          54          = 

hhd. 
1 

102      REDUCTION  OF  COMPOUND  NUMBERS. 

Note  1.  —  The  gallon  of  beer  measure  contains  282  cubic  inches;  and  has 
been  usually  reckoned,  36  gallons  equal  1  barrel;  2  hogsheads,  05/IO8  gallons, 
1  butt;  2  butts,  or  216  gallons,  1  tun. 

Note  2.  —  Beer  Measure  is  becoming  obsolete.  Milk  and  malt  liquors,  at 
the  present  time,  are  bought  and  sold,  very  generally,  by  wine  or  liquid 
measure. 

Examples. 

1.  How  many  pints  in  46khd.  49gal.? 

2.  In  20264  pints  how  many  hogsheads  ? 

3.  In  368hhd.  how  many  pints  ? 

4.  In  158976  pints  how  many  hogsheads  ? 

5.  At  29  cents  per  gallon,  what  cost  76  hogsheads  of  ale  ? 

Ans.  $1190.16. 

6.  How  much  may  be  obtained  by  selling  47hhd.  36gal.  of 
lager-bier  at  5  cents  a  quart?  Ans.  $514.80. 

DRY  MEASURE. 

140.     This    measure   is   used   in   measuring    grain,    fruit, 
salt,  &c. 

Table. 


2  Pints  (pt.) 
8  Quarts 
4  Pecks 

make 

lit 

u 

1  Quart, 
1  Peck 

1  Bushel, 

qt. 
pk 
bu. 

pts. 
8 



gal. 

1 

Pk. 

16 

= 

2 

=            1 

bu. 

64 

— 

8 

=             4 

= 

1 

Note  1.  —  The  Standard  Unit  of  Dry  Measure  adopted  by  the  United  States 
government  is  the  Winchester  bushel,  which  is  in  form  a  cylinder,  18£  inches 
in  diameter,  and  8  inches  deep,  containing  2150^  cubic  inches.  The  Stand- 
ard Imperial  bushel  of  Great  Britain  contains  22181^  cubic  inches,  so  that 
32  Imperial  bushels  equal  about  33  Winchester  bushels.  The  gallon  in  Dry 
Measure  contains  268|  cubic  inches. 

Note  2.  —  Of  wheat  a  standard  bushel  is  601b. ;  of  shelled  corn,  561b. ;  of 
corn  on  the  cob,  701b. ;  of  rye,  561b. ;  of  barley,  481b. ;  of  buckwheat  in  Pa., 
Ky.,  &c,  521b. ;  of  buckwheat  in  Mass.,  481b. ;  of  oats  in  Ohio,  111.,  Mass.,  &c, 
321b.;  of  oats  in  Ky.,  33£lb.;  of  oats  in  Me.,  301b.;  of  oats  in  Pa.,  241b.;  of 
clover-seed,  601b. ;  of  flax-seed,  561b. ;  of  Timothy-seed,  451b. ;  of  bran,  201b. ; 
of  beans,  601b. ;  of  onions,  in  Pa.,  Ky.,  &c,  571b. ;  of)  onions  in  Mass.,  52ib. ; 
of  salt  in  Ky.,  561b.;  of  salt  in  111.,  501b.;  of  dried  apples  in  Pa.,  221b.;  of 
dried  apples  in  111.,  241b. ;  of  dried  peaches  in  Pa.,  331b. ;  of  dried  peaches  in 
111.,  321b.;  of  stove  coal  in  111.,  801b.;  of  bituminous  coal  in  the  Western 
States,  761b. ;  and  of  hard-wood  charcoal,  301b.    The  weight  by  law,  of  a  few 


REDUCTION   OF   COMPOUND   NUMBERS.  103 

of  the  articles  named,  to  a  bushel,  is  not  uniform  in  all  the  States,  and  there- 
fore may  vary  slightly  from  the  above,  in  a  few  States  not  mentioned. 

Note  3.  —  In  some  places  it  is  customary,  in  measuring  coal,  potatoes,  and 
like  articles,  to  "  heap  "  the  bushel,  as  it  is  called,  and  in  that  case  5  even 
pecks  are  about  equal  to  1  "heaped  bushel."  The  "  coal  bushel,"  as  estab- 
lished by  laws  of  Massachusetts,  Ohio,  and  some  other  States,  is  of  greater  capa- 
city than  the  Winchester  bushel.  In  some  parts  of  the  United  States  a  chaldron, 
a  measure  of  coal,  consists  of  36  bushels ;  and  in  other  parts  of  the  country 
it  consists  of  32  bushels,  or  of  4  quarters,  each  consisting  of  8  bushels.  The 
quarter,  however,  in  England  is  8  Imperial  bushels,  a  measure  of  grain  equal 
to  5601b.,  or  one  quarter  of  a  ton  of  22401bs. 

Examples. 

1.  How  many  pints  in  35bu.  3pk.  ? 

2.  In  2288  pints  how  many  bushels  ? 

3.  In  676  chaldrons,  of  36  bushels  each,  how  many  pecks? 

4.  How  many  chaldrons,  of  36  bushels  each,  in  97344 
pecks  ? 

5.  A  grocer  purchased  50  bushels  of  potatoes,  by  "heaped" 
measure,  at  60  cents  a  bushel,  and  sold  the  same,  by  "  even  " 
measure,  at  15  cents  a  peck ;  did  he  gain  or  lose  by  the  oper- 
ation ?  Ans.   Gain  $  7.50. 

6.  If  I  purchase  by  weight,  in  Pennsylvania,  96  bushels  of 
oats,  at  42  cents  a  bushel,  and  sell  the  same  by  weight,  in  Ohio, 
at  45  cents  a  bushel,  shall  I  gain  or  lose  by  so  doing  ? 

Ans.   Lose  $  7.92. 

Dry,  Liquid,  and  Beer  Measures  Compared. 

141  •  The  relative  value  of  the  gallon  and  its  subdivisions, 
of  the  several  measures,  in  cubic  inches,  and  in  denominations 
of  each  other,  are  shown  in  the  following 

Table. 

cu.  in. 

1  gal.  B.  M.  =  282  =  lgal.  lpt.  3£gi.  L.  M.  =  lgal.  «pt.  D.  M. 
1  gal.  D.  M.  =  268§  ==  lgal.  lpt.  lifgi.  L.  M.  =  3qt.  ligpt.  B.  M. 
1  gal.  L.  M.  =  231  =  3qt.  £pt.  D.  M.  =  3qt.  |«pt.  B.  M. 

1  qt.    B.  M.  ==    10h  =  lqt.  Opt.  lffgi.  L.  M.    =  lqt.  ggt.  D.  M. 
1  qt.  D.  M.  =   67a  =  lqt.  IJJpt.  L.  M.  =  lpt.  3i|Jgi.  B.  M. 

1  qt.  L.  M.  =  57|  =  l||pt.  D.  M.  =  lfjpt.  B.  M. 

1  pt.  B.  M.  =  35.i  =  l||pt.  L.  M.  mm  1^-pt.  I).  M. 

1  pt.   D.  M.  =   333  ^  ipt.  ggj,  L.  M.  =  3||igi.  B.  M. 

1  pt.  L.  M.  =   28  j  =  Jjpt.  D.  M.  =  iipt.  B.  M. 

1  gi.  L.  M.  =     lJ-2  =  &pt  D.  M.  ==  $#t.  B.  M. 


104  REDUCTION   OP   COMPOUND   NUMBERS. 

Note  1.  —  By  the  table,  it  is  evident  that  each  of  the  measures  of  capacity- 
is  a  species  of  cubic  measure;  and  to  change  cubic  measure,  expressed  in 
cubic  inches,  to  any  denomination  of  either  dry,  liquid,  or  beer  measure,  divide 
by  the  number  of  cubic  inches  required  to  make  a  unit  of  the  proposed  denominar 
tion.  Thus,  to  reduce  14ft.  294in.  cubic  measure  to  gallons  of  liquid  measure: 
14  cu.  ft.  294  cu.  in.  =  24486  cu.  in. ;  24486  cu.  in. -f  231  =  106  gallons,  liquid 
measure. 

Note  2.  —  To  change  a  quantity  from  one  measure  of  capacity  to  its  equiv- 
alent in  another,  reduce  the  given  quantity  to  cubic  inches,  and  then  find  their 
value  in  denominations  of  the  proposed  measure. 

Examples. 

1.  Change  4hhd.  15gal.  beer  measure  to  liquid  measure  ? 

2.  Change  4hhd.  30gal.  liquid  measure  to  beer  measure  ? 

3.  If  a  milkman  buy  milk  at  4  cents  per  quart,  beer  meas- 
ure, and  sell  the  same  at  6  cents  per  quart,  liquid  measure, 
what  will  he  gain  in  disposing  of  2820  gallons  ? 

Ans.  $  375.02ff 

4.  If  2538  gallons  of  milk  have  been  purchased  by  liquid 
measure,  at  4  cents  per  quart,  and  the  same  has  been  sold 
by  beer  measure  at  G  cents  per  quart,  what  has  been  the 
gain?  Ans.  $92.88. 

5.  A  merchant  bought  385  bushels  of  seed  peas  at  $  4.00 
per  bushel,  dry  measure.  lie  sold  the  same  at  20  cents 
per  quart,  liquid  measure.  "What  did  he  gain  by  the  pur- 
chase? Ans.  $1327.20. 

6.  J.  Day  bought  1000  bushels  of  corn  at  $  1.05  per  bushel, 
dry  measure,  and  sold  the  same  at  $  1.12  per  bushel,  liquid 
measure.  Did  he  gain  or  lose  by  the  operation,  and  how 
much  ? 

7.  My  hogshead  contains  30  cubic  feet.  How  many  more 
gallons  of  dry  measure  will  it  contain,  than  of  beer  meas- 
ure? 

8.  Bought  of  my  neighbor,  John  Smith,  365  gallons  of  milk, 
at  5  cents  per  quart ;  but  by  mistake  he  measured  it  in  his 
liquid  measure.     How  much  did  I  lose  ? 

MEASUEE  OF   TIME. 

142.  This  measure  is  applied  to  the  various  divisions  and 
subdivisions  into  which  time  is  divided. 


REDUCTION   OF   COMPOUND   NUMBERS.  105 

Table. 


60    Seconds  (sec.)                         make 

1  Minute, 

m, 

60    Minutes                                        " 

1  Hour, 

h. 

24    Hours                                           " 

iDay, 

d. 

7   Days                                            " 

1  Week, 

w. 

365£  Days,  or  52  weeks  1%  days,        " 

1  Julian  Year, 

y- 

12    Calendar  Months  (mo.)             " 

1  Year. 

sec.                             m. 

60      ==                 1                    h. 

3600      =               60     =           1 

d. 

86400      =          1440     =         24     = 

1                   w. 

604800      =        10080     =       168     = 

=         7=1 

3 

31557600      =      525960     =     8766     = 

=     365i               = 

jr- 


Note  1.  —  The  true  Solar  or  Tropical  Year  is  the  time  measured  from  the 
sun's  leaving  either  equinox  or  solstice  to  its  return  to  the  same  again,  and  is 
365d.  5h.  48m.  49j70sec. 

The  Julian  Year,  so  called  from  the  calendar  instituted  by  Julius  Caesar, 
contains  365^  days,  as  a  medium;  three  years  in  succession  containing  365 
days,  and  the  fourth  year  366  days ;  which,  as  compared  with  the  true  solar 
year,  produces  an  average  yearly  error  of  11m.  lOj^sec,  or  a  difference  that 
would  amount  to  1  whole  day  in  about  120  years. 

The  Gregorian  Year,  or  that  instituted  by  Pope  Gregory  XIII.,  in  the  year 
1582,  and  which  is  now  the  Civil  or  Legal  Year  in  use  among  most  nations  of 
the  earth,  contains,  like  the  Julian  year,  365  days  for  three  years  in  succession, 
and  366  days  for  the  fourth,  excepting  the  last  year  of  the  odd  centuries.  The 
Gregorian  year  is  so  nearly  correct  as  to  err  only  1  day  in  3866  years,  a 
difference  so  little  as  hardly  to  be  worth  taking  into  account. 

The  manner  of  reckoning  time  according  to  the  Julian  Calendar  is  termed 
Old  Style,  and  that  according  to  the  reformed  calendar  of  Gregory,  New  Style. 
England  did  not  adopt  the  new  style  till  1752,  when.,  according  to  an  act  of 
Parliament,  the  difference  between  the  two  styles,  which  then  amounted  to  11 
days,  was  removed,  by  the  day  following  the  2d  of  September  of  that  year 
being  accounted  the  14th  day.  The  difference  now  between  old  and  new  style 
is  12  days. 

A  Common  Year  is  one  of  365  days,  and  a  Leap  or  Bissextile  Year  is  one  of 
366  days.  Any  year,  excepting  the  last  year  of  the  odd  centuries,  that  can 
be  divided  by  4  without  a  remainder,  is  Leap  Year. 

A  Sidereal  Year  is  the  time  in  which  the  earth  revolves  round  the  sun,  and 
is  365d.  6h.  9m.  9  6_sec. 

Note  2.  — The  names  of  the  12  calendar  months,  composing  the  civil  year, 
are  January,  February,  March,  April,  May,  June,  July,  August,  September, 
October,  November,  December,  and  the  number  of  days  in  each  may  be  readily 
remembered  by  the  following  lines :  — 

"  Thirty  days  hath  September, 
April,  June,  and  November ; 
And  all  the  rest  have  thirty-one, 
Save  February,  which  alone 
Hath  twenty -eight ;  and  this,  in  fine, 
One  year  in  four  hath  twenty-nine." 


106 


REDUCTION   OF   COMPOUND   NUMBERS. 


TABLE 

Showing  the  Number  of  Days  from  any  Day  of  one  Month  to  the 
same  Day  of  any  other  Month  in  the  same  Year. 


from  any 

DAY  OF 

to  the  same  day  of  next 

Jan.  Feb. 

Mar. 

Apr. 

May. 

June. 

July. 

Aug. 

Sept. 

Oct. 

Nov. 
304 

Dec. 
834 

January 

365 

31 

59 

90 

120 

151 

181 

212 

243 

273 

February 

334 

365 

28 

59 

89 

120 

150 

181 

212 

242 

273 

303 

March 

306 

337 

365 

31 

61 

92 

122 

153 

184 

214 

245 

275 

April 

275 

306 

334 

365 

30 

61 

91 

122 

153 

183 

214 

244 

May 

245 

267 

304 

335 

365 

31 

61 

92 

123 

153 

184 

214 

June 

214 

245 

273 

304 

334 

365 

30 

61 

92 

122 

153 

183 

July 

184 

215 

243 

274 

304 

335 

365 

31 

62 

92 

123 

153 

August 

153 

184 

212 

243 

273 

304 

334 

365 

31 

61 

92 

122 

September 

122 

153 

181 

212 

242 

273 

303 

334 

365 

30 

61 

91 

October 

92 

123 

151 

182 

212 

243 

273 

304 

335 

365 

31 

61 

November 

61 

92 

120 

151 

181 

212 

242 

273 

304 

334 

365 

30 

December 

31 

62 

90 

121 

151 

182 

212 

243 

274 

304 

335 

365 

For  example,  suppose  we  wish  to  find  the  number  of  days  from  April  4th  to 
November  4th,  we  look  for  April  in  the  left-hand  vertical  column,  and  for  No- 
vember at  the  top,  and  where  the  lines  intersect  is  214,  the  number  sought. 
Again,  if  we  wish  the  number  of  days  from  June  10th  to  September  16th,  we 
find  the  difference  between  June  10th  and  September  10th  to  be  92  days,  and 
add  6  clays  for  the  excess  of  the  16th  over  the  10th  of  September,  and  so  we 
have  98  days  as  the  exact  difference. 

If  the  end  of  February  be  included  between  the  points  of  time,  a  day  must 
be  added  in  leap  year. 

"When  the  time  includes  more  than  one  year,  there  must  be  added  365  days 
for  each  year. 

Examples. 

1.  How  many  seconds  in  a  solar  year  ? 

2.  In  31556929  seconds  how  many  days  ? 

3.  How  many  seconds  from  the  deluge,  which  took  place 
2348  years  before  Christ,  to  the  year  1856,  a  year  being  365£ 
days?  Aiis.  132636592800. 

4.  In  74726807872  seconds  how  many  solar  years  ? 

Ans.  2368  years. 

5.  How  long  did  the  last  war  with  England  continue ;  it 
having  commenced  June  18th,  1812,  and  ended  February 
17th,  1815  ?  Ans.  974  days  =  2  years  244  days. 


CIRCULAR  ANGULAR  MEASURE. 

143.  Circular  Angular  Measure  is  applied  to  the  measure- 
ment of  circles  and  angles,  and  is  used  in  reckoning  latitude  and 
longitude,  and  the  revolutions  of  the  planets  round  the  sun. 


REDUCTION    OP    COMPOUND    NUMBERS. 


107 


60  Seconds  (") 

make 

1  Minute,                                        '. 

GO  Minutes 

u 

1  Degree,                                        °. 

30  Degrees 

a 

1  Sign,                                            S. 

12  Signs,  or 

360  Degrees, 

The  Circle  of  the  Zodiac,            C. 

60 

— 

/ 
1 

o 

3600 

= 

60 

mm 

1                             S. 

108000 

= 

1800 

mm 

30          =            1                       C. 

1296000 

= 

21600 

= 

360         =         12         =         1 

Note  1.  —  A   Circle  is   a   plane  figure 

bounded  by  a  curve  line,  all  parts  of  which 
are  equally  distinct  from  a  point  called  its 
centre. 

The  Circumference  of  a  circle  is  the  line 
which  bounds  it,  as  shown  by  the  diagram. 
An  Arc  of  a  circle  is  any  part  of  its  cir- 
cumference; as  AB. 

A  Radius  of  a  circle  is  a  straight  line 
drawn  from  its  centre  to  its  circumference ; 
as  CA,  CB,  or  CD. 

Every  circle,  large  or  small,  is  supposed 
to  be  divided  into  360  equal  parts,  called 
degrees. 
A  Quadrant  is  one  fourth  of  a  circle,  or  an  arc  of  90° ;  as  AB. 
An  Angle,  as  ACB,  is  the  inclination  or  opening  of  two  lines  which  meet  at 
a  point,  as  C.    The  point  is  the  vertex  of  the  angle.    If  a  circle  be  drawn 
around  the  vertex  of  an  angle  as  a  centre,  the  two  sides  of  the  angle,  as  radii 
of  the  circle,  will  include  an  arc,  which  is  the  measure  of  the  angle ;  as  the 
arc  AD  =  120°  is  the  measure  of  the  angle  ACD,  and  AB  ==  90©,  the  meas- 
ure of  the  angle  ACB;   hence  the  one  is  called  an  angle  of  120°,  and  the 
other  an  angle  of  90°. 

Note  2.  —  As  the  earth  turns  on  its  axis  from  west  to  east  every  24  hours, 
the  sun  appears  to  pass  from  east  to  west  §5  of  360o  of  longitude  every  hour, 
or  15°  of  longitude  in  1  hbur's  time,  or  1°  in  4  minutes  of  time,  and  1'  in  4 
seconds  of  time ;  so  that,  for  instance,  when  it  is  noon  at  any  place,  it  is  1 
hour  earlier  for  every  15°  of  longitude  westward,  and  1  hour  later  for  every 
15°  of  longitude  eastward.  Thus,  Boston  being  71°  4'  west  of  Greenwich, 
and  San  Francisco  51°  17'  west  of  Boston,  when  it  is  noon  at  Boston,  it  is  4h. 
44m.  16sec.  past  noon  at  Greenwich,  and  wanting  3h.  25m.  8sec.  of  noon  at 
San  Francisco. 

Examples. 

1.  How  many  minutes  in  27S.  27°  43'? 

2.  In  50263  minutes  how  many  signs  ? 

3.  How  many  seconds  in  44S.  18°  57'  23"? 

4.  How  many  signs  in  4820243"? 

5.  How  many  seconds  in  360°  ? 

6.  How  many  degrees  in  1296000"? 


108 


REDUCTION   OF   COMPOUND   NUMBERS. 


MISCELLANEOUS   TABLE. 

144.     The  following  denominations,  frequently  used,  are  not 
embraced  in  the  preceding  tables. 


12   units 

] 

nake 

1  dozen. 

12   dozen 

u 

1  gross. 

1 2   gross 

II 

1  great  gross, 

20   units 

U 

1  score. 

24   sheets  of  Paper 

u 

1  quire. 

20   quires 

a 

1  ream. 

2  reams 

ll 

1  bundle. 

5   bundles 

u 

1  bale. 

14  pounds  of  Iron  or  Lead 

U 

1  stone. 

21£  stone 

K 

ipig- 

8   pigs 

II 

1  fother 

18   inches 

a 

1  cubit. 

24|  cubic  feet  of  Stone 

it, 

1  perch. 

60  pairs  of  Shoes 

M 

1  case. 

25   pounds  of  Powder 

II 

1  keg. 

56    pounds  of  Butter 

It 

1  firkin. 

100  pounds  of  Fish 

<( 

1  quintal. 

196   pounds  of  Flour 

II 

1  barrel. 

200  pounds  of  Beef 

M 

1  barrel. 

200   pounds  of  Pork 

u 

1  barrel. 

200  pounds  of  Shad  or  Salmon. 

,  in  N.  Y.  &  Ct. 

n 

1  barrel. 

220   pounds  of  Fish,  in  Md. 

u 

1  barrel. 

256   pounds  of  Soap 

II 

1  barrel. 

300  pounds  of  Hydraulic  Cement 

u 

1  barrel. 

30  gallons  of  Fish,  in  Mass. 

M 

1  barrel. 

5   bushels  of  Shelled  Corn,  in 

Southern  States 

1  " 

1  barrel. 

8   bushels  of  Salt 

U 

1  hogshead. 

Note  1.  —  A  sheet  of  paper  folded  into  2  leaves  forms  a  folio;  a  sheet  folded 
into  4  leaves,  a  quarto;  a  sheet  folded  into  8  leaves,  an  octavo;  a  sheet  folded 
into  12  leaves,  a  12mo;  a  sheet  folded  into  18  leaves,  an  18mo;  a  sheet  folded 
into  24  leaves,  a  24mo. 

Note  2.  —  Of  shoemaker's  measure,  No.  1  of  small  size  is  4£  inches  in 
length ;  and  No.  1,  large  size,  is  8££  inches  in  length ;  and  each  succeeding 
number  of  either  size  is  J  of  an  inch  additional  length. 


Examples. 

i.  j  naies  4  bundles  1  ream  10  quires  of  paper  how 

any  sheets  ?  Ans.  23760. 

2.  In  23760  sheets  of  paper  how  many  bales  ? 

3.  In  10  fothers  6  pigs  8  stones  of  iron  how  many  pounds  ? 

4.  In  25998  pounds  of  iron  how  many  fothers  ? 


1.  In  4  bales 
many  sheets  ? 


REDUCTION   OF   COMPOUND   NUMBERS.  109 

5.  At  23  cents  a  pound,  what  will  12  firkins  of  butter,  cost? 

6.  If  $  22  is  paid  for  a  barrel  of  pork,  how  much  is  that  by 
the  pound?  Ans.  11  cents. 

7.  How  much  must  be  paid  for  302  hogsheads  of  salt  at 
$  0.30  a  bushel  ?  Ans.  $  724.80. 

8.  At  $  4  per  quintal,  how  many  pounds  of  fish  may  be 
bought  for  $  50.24  ?  Ans.  1256  pounds. 

9.  If  the  wholesale  price  of  one  writing-book  be  4 J  cents, 
what  will  be  the  cost  of  a  great-gross  of  writing-books  ? 

10.  A  dairyman  sells  2  firkins  of  butter  at  20  cents  a  pound, 
and  takes  in  pay  half  a  barrel  of  flour  at  5  cents  a  pound,  and 
the  balance  in  cash.     How  much  cash  does  he  receive  ? 

Ans.  $  17.50. 

MISCELLANEOUS  EXAMPLES  IN  EEDUCTION. 

1.  In  57£.  15s.  how  many  half-pence  ?         .   Ans.  27720. 

2.  In  591b.  13pwt.  15gr.  how  many  grains  ? 

3.  In  340167  grains  how  many  pounds  ? 

4.  How  many  ells  English  in  761  yards  ? 

Ans.  608  E.  E.  4qr. 

5.  How  many  yards  in  61  ells  Flemish  ?     Ans.  45yd.  3qr. 

6.  How  many  bottles,  that  contain  3  pints  each,  will  it  take 
to  hold  a  hogshead  of  wine  ?  Ans.  168. 

7.  How  many  steps,  of  2ft.  8in.  each,  will  a  man  take  in 
walking  from  Bradford  to  Newburyport,  the  distance  being 
fifteen  miles  ?  Ans.  29700. 

8.  How  many  spoons,  each  weighing  2oz.  12pwt.,  can  be 
made  from  51b.  2oz.  8pwt.  of  silver  ?  Ans.  24. 

9.  How  many  times  will  the  wheel  of  a  coach  revolve,  whose 
circumference  is  14ft.  9in.,  in  passing  from  Boston  to  Washing- 
ton, the  distance  being  436  miles  ?  Ans.  156073T3T9T. 

10.  I  have  a  field  of  corn,  consisting  of  123  rows,  and  each 
row  contains  78  hills,  and  each  hill  has  4  ears  of  corn ;  now 
if  it  takes  8  ears  of  corn  to  make  a  quart,  how  many  bushels 
does  the  field  contain  ?  Ans.  149bu.  3pk.  5qt.  Opt. 

11.  If  it  take  5yd.  2qr.  3na.  to  make  a  suit  of  clothes,  how 
many  suits  can  be  made  from  182  yards  ? 

12.  A  goldsmith  wishes  to  make  a  number  of  rings,  each 

N     10  # 


110      REDUCTION  OP  COMPOUND  NUMBERS. 

weighing  5pwt.  10gr.,  from  31b.  loz.  2pwt.  2gr.  of  gold;  how 
many  will  there  be  ?  Ans.  137. 

13.  How  many  shingles  will  it  take  to  cover  the  roof  of  a 
building,  whifth  is  60  feet  long  and  56  feet  wide,  allowing  each 
shingle  to  be  4  inches  wide  and  18  inches  long,  and  to  lie  one 
third  to  the  weather  ?  Ans.  20160. 

14.  There  is  a  house  56  feet  long,  and  each  of  the  two  sides 
of  the  roof  is  25  feet  wide  ;  how  many  shingles  will  it  take  to 
cover  it,  if  it  require  6  shingles  to  cover  a  square  foot  ? 

Ans.  16800. 

15.  If  a  man  can  travel  22m.  3fur.  17rd.  a  day,  how  long 
would  it  take  him  to  walk  round  the  globe,  the  distance  being 
about  25000  miles  ?  Ans.  1114fff f  days. 

16.  If  a  family  consume  71b.  lOoz.  of  sugar  in  a  week,  how 
long  would  lOcwt.  3qr.  161b.  last  them  ?     Ans.  143^5T  weeks. 

17.  What  will  ^  hogsheads  of  wine  cost,  at  9  cents  a  quart? 

18.  What  will  15  hogsheads  of  beer  cost,  at  3  cents  a 
pint?  Ans.  $194.40. 

19.  What  will  73  bushels  of  meal  cost,  at  2  cents  a 
quart?  Ans.  $46.72. 

20.  A  merchant  has  29  bales  of  cotton  cloth ;  each  bale 
contains  57  yards  ;  what  is  the  value  of  the  whole  at  15  cents 
a  yard  ?  Ans.  $  247.95. 

21.  There  is  a  certain  pile  of  wood  120  feet  long,  4£ 
feet  high,  and  4  feet  wide ;  what  is  its  value  at  $  4.00  per 
cord?  Ans.  $67.50. 

22.  How  much  must  be  paid,  at  twenty  cents  a  square  yard, 
for  plastering  overhead  in  a  room  which  is  33  feet  long  and  18 
feet  wide  ? 

23.  An  apothecary,  in  compounding  20  boxes  of  pills,  each 
box  containing  25  pills,  used  6  grains  of  aloes,  5  grains  of 
rhubarb,  and  4  grains  of  calomel  in  each  pill.  What  was  the 
entire  quantity  used  ?  Ans.  7500  grains. 

24.  Purchased  a  cargo  of  molasses,  consisting  of  87  hogs- 
heads ;  wTiat  is  the  value  of  it  at  33  cents  a  gallon  ? 

Ans.  $  1808.73. 

25.  If  a  cubic  foot  of  white-oak  wood  weighs  880  ounces, 
and  a  cubic  foot  of  white-pine  wood  weighs  480  ounces,  how 
much  will  a  load  weigh,  which  is  composed  of  half  a  cord  of 
white-oak,  anil  of  a  cord  of  white-pine  ?  Ans.  73601bs. 


ADDITION   OF   COMPOUND   NUMBERS.  Ill 

26.  How  many  chests  of  tea,  weighing  24  pounds,  at  43 
cents  a  pound,  can  be  bought  for  $  1548  ? 

27.  Joseph  Eldredge  received  $  10,  as  a  birthday  present, 
from  his  father,  on  every  29th  day  of  February,  from  1837  to 
1857.     How  much  less  than  $  200  did  he  receive,  in  all  ? 

Ans.  $  150. 

28.  If  25f  grains  of  standard  gold  be  worth  $1,  how 
many  pounds  avoirdupois  of  standard  gold  will  be  worth 
$  1,000,000  ?  Ans.  3685f  pounds. 

29.  A  merchant,  who  had  bought  188  gallons  of  molasses, 
at  40  cents  a  gallon,  intended  to  have  it  sold  at  the  rate  of  50 
cents  a  gallon ;  but  his  shop-boy  retailed  half  of  the  quan- 
tity at  12^  cents  a  quart,  beer  measure,  when,  finding  he  had 
made  a  blunder,  he  sold  the  balance  at  14  cents  a  quart,  wine 
measure,  thereby  expecting  to  exactly  make  up  for  the  mis- 
take.    How  much  less  d^l  the  whole  bring  than  was  intended  ? 

Ans.  $  2.86. 


ADDITION  OF   COMPOUND  NUMBERS. 

145i  Addition  of  Compound  Numbers  is  the  process  of 
finding  the  amount  of  two  or  more  compound  numbers. 

Ex.  1.  Required  the  amount  of  31£.  17s.  9d.  2far.;  16£. 
16s.  6d.  lfar.;  16£.  lis.  lid.  lfar.;  19£.  19s.  9d.  3far. ;  61£. 
17s.  Id.  2far.  Ans.  147£.  3s.  2d.  lfar. 

Having  written  units  of  the  same 
denomination  in  the  same  column,  we 
find  the  sum  of  farthings  in  the  right- 
hand  column  to  be  9  farthings  =  2d. 
lfar.     We  write  the  lfar.  under  the 
column  of  farthings,  and  carry  the  2d. 
to  the  column  of  pence ;  the  sum  of 
which   is  38d.  =  3s.  2d.     We  write 
the  2d.  under  the  column  of  pence, 
and   carry  the  3s.  to  the  column  of 
shillings  ;  the  sum  of  which  is  83s.  =  4£.  3s.     Having  written  the 
3s.  under  the  column  of  shillings,  we  carry  the  4£.  to  the  column  of 
pounds,  and  find  the  entire  amount  sought  to  be  14 7£.  3s.  2d.  lfar. 

The  same  result  can  be  arrived  at  by  reducing  the  numbers  as  they 
are*  added  in  their  respective  columns.  Thus,  beginning  with  far- 
things, we  can  add,  in  this  way :  2far.  -f-  3far.  =  5far.  =  Id.  lfar. ; 


£. 

OPERATION. 

s.           d. 

far. 

31 

17 

9 

2 

16 

16 

6 

1 

16 

11 

11 

1 

19 

19 

9 

3 

61 

17 

1 

2 

Ans.  14  7 

3 

2 

1 

112  ADDITION   OP  COMPOUND  NUMBERS. 

§ 

and  lfar.  =-  Id.  2far.,  and  lfar.  =  Id.  3far.,  and  2far.  =  2d.  lfar. 
Writing  the  lfar.  under  the  column  of  farthings,  we  carry  the  2d.  to 
the  column  of  pence ;  and  add,  2d.  (carried)  -J-  1  =  3d.,  and  9d.  = 
12d.  =  Is.,  and  lid.  =  Is.  lid.,  and  6d.  =  2s.  5d.,  and  9d.  =  3s. 
2d.  Writing  the  2d.  under  the  column  of  pence,  we  carry  the  3s.  to 
the  column  of  shillings ;  and  add,  3s.  (carried)  -[-  17s.  =  20s.  =  l£., 
and  19s.  =  l£.  19s.,  and  lis.  =  2£.  10s.,  and  16s.  =  3£.  6s.,  and 
17s.  =  4£.  3s.  Writing  the  3s.  under  the  column  of  shillings,  we 
carry  the  4£.  to  the  column  of  pounds,  and  so  find  the  whole  amount 
to  be,  as  before,  14 7£.  3s.  2d.  lfar. 

The  last  method  of  operation  may  be  rendered  more  concise,  as  it 
should  always  be  in  practice,  by  merely  naming  results  as  the  adding 
is  performed  (Art.  45). 

From  the  illustrations  given,  it  is  evident  that  the  adding  of  com- 
pound numbers  is  like  that  of  simple  numbers,  except  in  carrying ; 
which  difference  holds  also  in  subtracting,  multiplying,  and  dividing 
compound  numbers. 

Rule. — Write  all  the  given  numbers,  so  that  units  of  the  same  de- 
nomination may  stand  in  the  same  column. 

Add  as  in  addition  of  simple  numbers  p  and  carry,  from  column  to 
column,  one  for  as  many  units  as  it  takes  of  the  denomination  added  to 
make  a  unit  of  the  denomination  next  higher. 

Proof.  —  The  proof  is  the  same  as  in  addition  of  simple 
numbers. 

Examples. 
2.  3. 

Ton.      cwt.      qr.       lb.  oz.         dr.  cwt.      qr.       lb.         oz.        dr. 

61193171515  612111114 

63     133161111  163151511 

51     12     3     17        7        6  41     3     13        9        9 

61161111212  38     2111010 

13     13     3     12     13     15  42     1        9        8     13 

71182131414  313171112 


324  15 

2 
4, 

15 

12   9 

5. 

lb. 

oz. 

pwt 

gr. 

lb. 

oz. 

pwt. 

gr. 

16 

11 

19 

23 

123 

9 

7 

13 

31 

10 

18 

16 

98 

11 

17 

14 

63 

9 

12 

15 

49 

7 

13 

21 

17 

8 

13 

12 

13 

10 

10 

20 

61 

7 

12 

16   ' 

47 

9 

19 

23 

17 

6 

17 

22 

51 

5 

15 

15 



209        7     15 


ADDITION   OP   COMPOUND   NUMBERS.  113 


6. 

7. 

lb 

5 

5 

B 

gr. 

lb 

§ 

5 

3       gr. 

27 

1  1 

7 

2 

19 

37 

9 

6 

1  18 

16 

10 

6 

1 

13 

14 

4 

4 

2  11 

41 

9 

3 

2 

16 

61 

6 

3 

2   6 

38 

10 

5 

2 

14 

41 

4 

7 

2  16 

41 

4 

4 

1 

11 

39 

8 

4 

1  12 

16 

6 

6 

2 

6 

51 

11 

7 

2  19 

183 

6 

3 

8. 

1 

19 

9. 

eg..   m. 

fur. 

rd. 

ft.    in. 

m.   fur.   rd.   j 

rd.  ft.   3 

17  69  7  39  16  11  69  7  31  5  2  11 

6162  31712  9  1661641  6 

1616  61613  10  617  32  32  10 

48  19  3  15  15  9  73  3  16  4  2  9 


17 
33 

58   6 
35   5 

33 
19 

14   7 
9   9 

19  4  14  1  1 

75  5  25  5  2 

8 

7 

195 

55^     1 
i-4 

24 

i    7 

195  55   5  24   1   1 

Note.  —  As  half  a  mile  is  equal  to  4  furlongs,  we  add  them  to  the  1  fur- 
long, which  makes  5  furlongs.  And  as  half  a  foot  is  equal  to  6  inches,  we  add 
them  to  the  7  inches,  which  makes  13  inches ;  and  these  are  equal  to  1  foot 
1  inch. 

10.  Add  together  37yd.  3qr.  3na.  2in. ;  61yd.  3qr.  Ina.  lin.; 
13yd.  2qr.  2na.  2in.;  32yd.  lqr.  Ina.  lin.;  61yd.  2qr.  2na. 
2in.;  and  22yd.  lqr.  3na.  Ans.  229yd.  3qr.  3na.  l£in. 

11.  Add  together  671E.E.  lqr.  Ina.  lin.;  161E.E.  3qr.  3na. 
2in. ;  617E.E.  3qr.  Ina.  2in. ;  178E.E.  3qr.  2na.  lin. ;  717E.E. 
2qr.  Ina.  2in. ;  and  166yd.  3qr.  2na.  lin. 

12.  Add  together  761A.  3R.  37p.  260ft.  125in.;  131A.  2R. 
16p.  135ft.  112in.;  613A.  1R.  14p.  116ft.  131in.;  161A.  3R. 
13p.  116ft.  123in.;  321A.  2R.  31p.  97ft.  96in.;  and  47 A.  3R. 
19p.  91ft.  48in.  Ans.  2038A.  1R.  13p.  2ft.  95in. 

13.  Add  together  38A.  1R.  39p.  272ft.;  61A.  3R.  38p.- 
167ft. ;  35A.  3R.  19p.  198ft.;  47 A.  3R.  16p.  271ft.;  86A. 
2R.  13p.  198ft.;  and  46A.  1R.  14p.  269ft. 

14.  Add  together  17m.  7fur.  9ch.  3p.  241.;  16m.  3fur.  4ch. 

10* 


114  ADDITION   OP   COMPOUND   NUMBERS. 

lp.  151. ;  27m.  4fur.  6ch.  2p.  171.;  18m.  6fur.  3ch.  3p.  211..; 
61m.  7fur.  7ch.  2p.  161.;  and  17m.  lfur.  8ch.  2p.  191. 

Ans.  160m.  Ofur.  lch.  lp.  121. 

15.  Required  the  sum  of  27m.  4fur.  3ch.  lp.  211.;  29m. 
3fur.  lch.  3p.  231.;  67m.  3fur.  3ch.  lp.  191.;  21m.  7fur. 
lch.  3p.  161.;  16m.  7fur.  9ch.  3p.  131.;  and  31m.  4fur.  8ch. 
lp.  201. 

16.  Required  the  value  of  29T.  36ft.  1279in.  +  69T.  19ft. 
1345in.  +  67T.  18ft.  1099in.  +  71T.  14ft.  1727in.  +  43T. 
35ft.  916in.  +  53T.  17ft.  1719m. 

Ans.  335T.  23ft.  1173in. 

17.  Add  together  61C.  127ft.  1161in.;  37C.  89ft.  1711in.; 
61C.  98ft.  1336in.;  43C.  56ft.  1678m.  91T.  119ft.  1357in.  and 
81C.  115ft.  1129in. 

18.  Required  the  value  of  61  tuns  3hhd.  62gal.  3qt.  lpt.  -[- 
39  tuns  2hhd.  16gal.  lqt.  lpt.  +  68  tuns  3hhd.57gal.  2qt.  lpt. 
-f-  87  tuns  3hhd.  45gal.  3qt.  lpt.  -f-  47  tuns  2hhd.  59gal.  3qt. 
lpt.  +  47  tuns  3hhd.  39gal.  2qt.  lpt. 

Ans.  354  tuns  Ohhd.  30gal.  lqt. 

19.  Required  the  value  of  67hhd.  15gal.  3qt.  lpt.  -f  16hhd. 
16gal.  3qt.  +  39hhd.  16gal.  3qt.  -f  47hhd.  62gal.  lqt.  lpt.  -f 
43hhd.  57gal.  3qt.  +  71hhd.  61gal.  3qt.  lpt. 

20.  Add  together  of  beer  measure  161hhd.  53gal.  3qt.  lpt. ; 
371hhd.  52gal.  3qt.  lpt.;  98hhd.  19gal.  lqt.;  47hhd.  43gal. 
lqt.  Opt.;  61hhd.  43gal.  lqt.  lpt.;  and  42hhd.  27gal.  3qt.  lpt. 

21.  Find  the  amount  of  37bu.  3pk.  5qt.  lpt.  -|-  61bu.  2pk. 
7qt.  lpt.  -f-  32bu.  3pk.  2qt.  +  71bu.  lpk.  6qt.  lpt.  +  61bu. 
lpk.  3qt.  lpt.  -j-  32bu.  3pk.  3qt.  lpt. 

Ans.  298bu.  Opk.  4qt.  lpt. 

22.  Required  the  value  of  31bu.  3pk.  3qt.  +  31bu.  3pk. 
lqt.  +  16bu.  3pk.  lqt.  +  15bu.  3pk.  -f  17bu.  3pk.  lqt.  + 
14bu.  3pk.  lqt. 

23.  Required  the  value  of  57y.  llmo.  27d.  23h.  29m.  55s. 
+  31y.  llmo.  18d.  19h.  19m.  39s.  +  46y.  9mo.  23d.  17h. 
28m.  56s.  +  43y.  lOmo.  16d.  18h.  17m.  48s.  +  32y.  9mo. 
19d.  16h.  23m.  28s.  +  14y.  lmo.  29d.  21h.  28m.  16s. 

Ans.  227y.  7mo.  16d.  21h.  28m.  2s. 

24.  Required  the  value  of  23w.  6d.  23h.  59m.  58s.  -f  51w. 
3d.  18h.  51m.  17s.  +  29w.  5d.  21h.  47m.  49s.  +  28w.  4d. 


ADDITION   OF    COMPOUND   NUMBERS.  115 

23h.  56m.  18s.  +  19w.  6d.  lOh.  18m.  53s.  -f  86w.  Id.  20h. 
40m.  51s. 

25.  Add  together  4S.  29°  59'  59";  6S.  17°  17'  29";  US. 
16°  W  58";  9S.  13°  46'  51";  5S.  27°  16'  42";  and  2S.  25° 
11'  17".  Ans.  5S.  10°  35'  16". 

26.  Add  together  US.  11°  16'  51";  6S.  6°  6'  16";  9S.  14° 
56'  56";  3S.  29°  29'  49";  9S.  17°  18'  58";  and  6S.  13°  13' 
52". 

Note.  —  We  divide  the  sum  of  the  signs,  in  the  last  two  questions,  by  12, 
and  write  down  the  remainder  only.  Since  the  circumference  of  a  circle  can- 
not exceed  12  signs  (Art.  143). 

27.  Bought  of  a  London  tailor  a  vest  for  1£.  13s.  4d.,  a  coat 
for  7£.  12s.  9d.,  pantaloons  for  2£.  3s.  9d.,  and  surtout  for 
9£.  8s.  Od. ;  what  was  the  whole  amount  ? 

Ans.  20£.  17s.  lOd. 

28.  Bought  a  silver  tankard,  weighing  lib.  8oz.  17pwt.  14gr., 
a  silver  can,  weighing  lib.  2oz.  12pwt,  a  porringer,  weighing 
lloz.  19pwt.  20gr. ;  and  3  dozen  of  spoons,  weighing  lib.  9oz. 
15pwt.  lOgr. ;  what  was  the  whole  weight? 

Ans.  51b.  9oz.  4pwt.  20  gr. 

29.  What  is  the  weight  of  a  mixture  of  31b  4§  25  23 
14gr.  of  aloe,  2tb  7§  65  19  13gr.  of  picra,  and  lit)  105  15 
23  17gr.  of  saffron?  Ans.  71b  10 §  35  19  4  gr. 

30.  Sold  4  loads  of  hay ;  the  first  weighed  27cwt.  3qr.  181b. ; 
the  second,  31cwt.  lqr.  151b. ;  the  third,  19cwt.  lqr.  151b. ; 
and  the  fourth,  38cwt.  2qr.  241b. ;  what  is  the  weight  of  the 
whole  ? 

31.  Bought  5  pieces  of  broadcloth;  the  first  contained  17yd. 
3qr.  2na. ;  second,  13yd.  2qr.  lna. ;  the  third,  87yd.  lqr.  3na. ; 
the  fourth,  27yd.  lqr.  2na.;  and  the  fifth,  29yd.  lqr.  2na. ; 
what  was  the  whole  quantity  purchased  ? 

Ans.  175yd.  2qr.  2na. 

32.  A  pedestrian  travelled,  the  first  week,  371m.  3fur.  37rd. 
5yd.  2ft.  10in.;  the  second  week,  289m.  2fur.  18rd.  3yd.  1ft. 
9in. ;  and  the  third  week,  399m.  7fur.  3ft.  llin. ;  how  many 
miles  did  he  travel?  Ans.  1060m.  5fur.  16rd.  5yd.  1ft. 

33.  A  man  has  three  farms;  the  first  contains  186A.  3R. 
14p. ;  the  second,  286A.  17p.;  and  the  third,  115 A.  2R. ;  how 
much  do  they  all  contain?  Ans.  588 A.  1R.  31p. 


116  SUBTRACTION   OF   COMPOUND   NUMBERS. 

34.  The  Moon  is  5S.  18°  14'  17"  east  of  the  Sun;  Jupiter  is 
7S.  10°  29'  28"  east  of  the  Moon;  Mars  is  US.  12°  ll7  56" 
east  of  Jupiter;  and  Herschel  is  78.  18°  38'  15"  east  of  Mars  ; 
how  far  is  Herschel  east  from  the  Sun  ? 

Ans.  7S.  29°  33'  56". 


OPERATION. 

£                    8. 

d. 

Min. 

617     11 

8 

Sub. 

181     15 

5 

SUBTRACTION  OF  COMPOUND  NUMBERS. 

146.     Subtraction  of  Compound  Numbers  is  the  process 
of  finding  the  difference  between  two  compound  numbers. 
Ex.  1.     From  617£.  lis.  8d.  take  181£.  15s.  5d. 

Ans.  435£.  16s.  3d. 

Having  placed  the  less  number  un- 
der the  greater,  pence  under  pence, 
shillings   under  shillings,   &c,   we  be- 
gin with  pence,  thus:    5d.   from   8d. 
"R  A  o  c     T~7     o~       leaves  3d.,   which  we  set  under  the 

Kem.    4  o  o     lb     o       column  of  pence.     As  we  cannot  take 

15s.  from  lis.,  we  add  20s.  =  l£.  to 
the  lis.,  making  31s.,  and  then  subtract  the  15s.  from  it,  and  set  the 
remainder,  16s.,  under  the  column  of  shillings.  Then,  having  added 
l£.  =  20s.  to  the  18l£.,  to  compensate  for  the  20s.  added  to  the  lis. 
in  the  minuend,  we  subtract  the  pounds  as  in  subtraction  of  simple 
numbers,  and  obtain  435£.  for  the  remainder,  and  as  the  result  com- 
plete, 435£.  16s.  3d. 

Rule.  —  Write  the  less  compound  number  under  the  greater,  so  that 
units  of  the  same  denomination  shall  stand  in  the  same  column. 

Subtract  as  in  subtraction  of  simple  numbers. 

If  any  number  in  the  subtrahend  is  larger  than  that  above  it,  add  to 
the  upper  number  as  many  units  as  make  one  of  the  next  higher  denomi- 
nation before  subtracting,  and  carry  one  to  the  next  lower  number  before 
subtracting  it. 

Proof.  —  The  proof  is  the  same  as  in  simple  subtraction. 
Examples. 


£ 

s.          d. 

87 

16     3J 

19 

17     9  i 

£ 

s. 

d. 

617 

1 1 

H 

181 

15 

8* 

67     18     5f 


SUBTRACTION   OF   COMPOUND  NUMBERS.  117 

4.  5. 

T.       cwt.      qr.       lb.      oz.       dr.  cwt.     qr.       lb.         oz. 

71     18     1     13     1     13  73     1     15     13 

19192168        5  19119     15 


51  : 

L8  2  21 

6. 

9   8 

7. 

lb. 

71 

oz.   pwt. 

3  12 

gr. 

15 

lb. 

58 

oz.  pwt, 

5  12 

10 

- 

16 

10  17 

20 

19 

9  17 

21 

54 

4  14 

8. 

19 

9. 

lb 
71 

§  3  3 
1  3  1 

gr. 

14 

lb 
15 

§39 

2  2  0 

gr- 

15 

18 

6  7  2 

19 

9 

9  1  1 

18 

52 

6  3  1 
10. 

15 

11. 

m.  fur. 

16  7 

rd.    ft. 

18   3 

in. 
1 

deg. 

38 

m. 

41 

fur.   rd. 

3  29 

rd.  ft. 
2  1 

in. 

7 

9  7 

19  16 

8 

29 

36 

5  31 

3  1 

9 

6    7     38       2|      5 

6     7     38        2      11 

Note.  —  As  half  a  foot  is  equal  to  6  inches,  we  add  them  to  the  5  inches, 
which  make  11  inches. 

12.  From  67yd.  lqr.  lna.  lin.  take  18yd.  2qr.  2na.  2in. 

Ans.  48yd.  2qr.  2na.  l£in. 

13.  From  51E.E.  2qr.  3na.  take  19E.E.  3qr.  lna. 

14.  From  56A.  1R.  19p.  119ft.  HOin.  take  17A.  3R.  13p. 
127ft.  113in.  Ans.  38A.  2R.  5p.  264ft.  33in. 

15.  Find  the  value  of  13A.  1R.  15p.  19yd.  1ft.  17in.  —  9A. 
3R.  16p.  30yd.  5ft.  17in. 

16.  Subtract  19m.  2fur.  lch.  3p.  211.  from  21m.  lfur.  3ch. 
2p.  191.  Ans.  lm.  7fur.  lch.  2p.  231. 

17.  From  28m.  6fur.  lch.  2p.  181.  take  15m.  7fur.  3ch.  lp. 
191. 

18.  Required  the  value  of  49T.  13ft.  1611in.  —  18T.  15ft. 
1719in.  Ans.  30T.  37ft.  1620in. 


118  SUBTRACTION   OF   COMPOUND  NUMBERS. 

19.  Required  the  value  of  361C.  47ft.  1178in.  —  197C. 
121ft.  1617in. 

20.  Subtract  lltun  lhhd.  28gal.  2qt.  lpt.  from  79tun  3hhd. 
19gal.  lqt.  lpt.  Ans.  68tun  lhhd.  53  gal.  3qt. 

21.  Subtract  of  beer  measure  191hhd.  19gal.  3qt.  from 
769hhd.  18gal.  lqt. 

22.  From  56ch.  2bu.  lpk.  take  38ch.  3bu.  lpk. 

Ans.  17ch.  35bu. 

23.  Required  the  value  of  25bu.  3pk.  lqt.  —  12bu.  3pk.  5qt. 

Ans.  12bu.  3pk.  4qt. 

24.  Required  the  difference  between  6mo.  16d.  13h.  27m. 
19s.  and  lmo.  22d.  16h.  41m.  37s. 

Ans.  4mo.  23d.  20h.  45m.  42s. 

25.  From  48y.  Omo.  15d.  19h.  27m.  31s.  take  19y.  lOmo. 
29d.  21h.  38m.  56s. 

26.  From  6S.  11°  12'  48"  subtract  9S.  8°  15'  56". 

Ans.  9S.  2°  56'  52". 

27.  Take  IS.  22°  19'  28"  from  4S.  19°  41'  22". 

28.  I  have  73A.  of  land ;  if  I  should  sell  5A.  3R.  lp.  7ft., 
how  much  should  I  have  left  ?     Ans.  67A.  OR.  38p.  265Jft. 

29.  A  owes  B  100£.  ;  what  will  remain  due  after  he  has 
paid  him  3s.  6£d.  ?  Ans.  99£.  16s.  5£d. 

30.  It  is  about  25,000  miles  round  the  globe  ;  if  a  man  shall 
have  travelled  43m.  17rd.  9in.,  how  much  will  remain  to  be 
travelled?  Ans.  24,956m.  7fur.  22rd.  15ft.  9in. 

31.  Bought  7  cords  of  wood;  2  cords  78ft.  having  been 
stolen,  how  much  remained  ? 

32.  I  have  15  yards  of  cloth ;  having  sold  3yd.  2qr.  lna., 
what  remains?  Ans.  11yd.  lqr.  3na. 

33.  If  a  wagon  loaded  with  hay  weighs  43cwt.  2qr.  181b., 
and  the  wagon  is  afterwards  found  to  weigh  9cwt.  3qr.  231b., 
what  is  the  weight  of  the  hay  ?  Ans.  33cwt.  2qr.  201b. 

34.  Bought  a  hogshead  of  wine,  and  by  an  accident  8gal. 
3qt.  lpt.  leaked  out ;  what  remains  ? 

35.  I  had  10A.  3R.  lOp.  of  land ;  and  I  have  sold  two  house- 
lots,  one  containing  1A.  2R.  13p.,  the  other  2 A.  2R.  op.;  how 
much  have  I  remaining  ?  Ans.  6 A.  2R.  32p. 

36.  The  Moon  moves  13°  10'  35"  in  a  solar  day,  and  the 
Sun  59'   8"  ;   now   supposing  them   both    to    start  from  the 


MULTIPLICATION   OP   COMPOUND   NUMBERS.  119 

same  point  in  the  heavens,  how  far  will  the  Moon  have  gained 
on  the  Sun  in  24  hours  ?  Ans.  12°  11'  27". 

37.  A  farmer  raised  136bu.  of  wheat;  if  he  sells  49bu.  2pk. 
7qt.  lpt.,  how  much  has  he  remaining  ? 

Ans.  86bu.  lpk.  Oqt.  lpt. 

38.  If  from  a  stick  of  timber  containing  2T.  18ft.  1410in. 
there  be  taken  38ft.  1720in.,  how  much  will  be  left  ? 

Ans.  IT.  19ft.  1418in. 


MULTIPLICATION  OF  COMPOUND  NUMBERS. 

147.  Multiplication  of  Compound  Numbers  is  the  pro- 
cess of  taking  a  compound  number  any  proposed  number  of 
times. 

Ex.  1.  "What  will  6  bales  of  cloth  cost,  at  7£.  12s.  7d.  per 
bale  ?  Ans.  45£.  15s.  6d. 

operation.  Having  written  the  multiplier  under 

*      s'       Jjj     the  lowest  denomination  of  the  multi- 
Multiplicand     7      12     7     piican(i,   we  multiply   thus :    7d.  X  6 

Multiplier      6     =  42d.  =  3s.  6d.     We  write  the  6d. 

Product  4~5     3~5     6     under  the  number  multiplied,  and  re- 

serve the  3s.  to  be  added  to  the  pro- 
duct of  the  shillings.  Then,  12s.  X  6  =  72s.,  and  3s.  (carried)  = 
75s.  =  3£.  15s.  We  write  the  15s.  under  the  column  of  shillings, 
and  reserve  the  3£.  to  be  added  to  the  product  of  the  pounds. 
Again,  7£.  X  6  =  42£.,  and  3£.  (carried)  =  4 5£.  This,  placed 
under  the  column  of  pounds,  gives  45£.  15s.  6d. 

Rule.  —  Multiply  each  denomination  of  the  compound  number  as  in 
multiplication  of  simple  numbers,  and  carry  as  in  addition  of  compound 
numbers. 

Proof  —  Write  down  by  themselves  the  several  products 
obtained  by  multiplying  each  denomination  of  the  multiplicand 
by  the  multiplier,  and  these  partial  products  added  together  will 
equal  the  entire  product,  if  the  work  be  right.     (Art.  60.) 

Note.  —  Going  a  second  time  carefully  over  the  work  is  a  good  way  of 
testing  its  accuracy.  On  learning  Division  of  Compound  Numbers,  the  pupil 
will  find  that  rule  a  better  method  of  proving  multiplication  of  compound 
numbers. 


120  MULTIPLICATION   OF   COMPOUND  NUMBERS. 

Examples. 
2.  Multiply  1£.  8s.  7d.  2far.  by  7.      Ans.  10£.  0s.  4d.  2far. 

PROOF    BY    ADDITION. 

9£d.  x  8     =  76d.  =    0£.  6s.  4d. 

8s.     x  8     =  64s.   =    3£.  4s.  Od. 

Ans.    19     10     4         2£.    x  8     =  16£.  =  16£.  0s.  Od. 

2£.  8s.  9id.  x  8  =  19£.  10s.  4d. 

Note.  —  The  answers  to  the  following  examples  may  be  found  in  corre- 
sponding numbers  of  examples  in  Division  of  Compound  Numbers. 

3.  4. 

T.   cwt  qr.   lb.    oz.    dr.  lb.   oz.   pwt.   gr. 

Gl  19  3  17  15  15        7  11  14  15 

9  5 


OPERATION 

£.              8. 

d. 

2                 8 

n 

8 

5.  6. 

lb.   oz.  pwt.   gr.  Jb    5    5  3       gr- 

32  8  17  12  38  10  5  2  14 

8  11 


7.  8. 

deg.   m.   fur.  rd.    ft.   in.  m.   fur.  ch.  p.    1. 

71  38  2  13  14  4  17  7  9  3  23 

12  12 


9.  Multiply  16A.  2R.  4p.  19yd.  7ft.  79in.  by  11. 

10.  Multiply  10yd.  3qr.  3na.  by  5. 

11.  Multiply  17tun  2hhd.  50gal.  lqt.  by  7. 

12.  Multiply  29hhd.  61gal.  3qt.  lpt.  3gi.  by  7. 

13.  Multiply  19bu.  2pk.  7qt.  lpt.  by  6. 

14.  What  is  the  value  of  13y.  31 6d.  loh.  27m.  39s.  X  8  ? 

15.  Multiply  16deg.  39m.  3fur.  39rd.  5yd.  2ft.  by  9. 

16.  If  a  man  gives  each  of  his  9  sons  23A.  3R.  19^p.,  what 
do  they  all  receive  ? 

17.  If  12  men  perform  a  piece  of  labor  in  7h.  24m.  30s., 
how  long  would  it  take  1  man  to  perform  the  same  task  ? 

18.  If  1  bag  contain  3bu.  2pk.  4qt.,  what  quantity  do  8  bags 
contain  ? 


OPERATION. 

T.        cwt.    qr. 

2        12 

lb. 

7 

weight  of  1  load. 

14     10     3 

17  = 
5 

weight  of  7  loads. 

MULTIPLICATION   OP    COMPOUND   NUMBEKS.  121 

148.     When  the  multiplier  is  a  composite  number,  and  none 
of  its  factors  exceed  12. 

Ex.  1.  What  will  35  loads  of  coal  weigh,  if  each  load  weighs 
2T.  lcwt.  2qr.  61b.  ? 

We  find  the  num- 
ber 35  equal  to  the 
product  of  7  and  5  ; 
we  therefore  multiply 
the  weight  of  1  load 
by  7,  and  then  that 
product  by  5  ;  and  the 

72  14  2  10  =  weight  of  35  loads,  last  product  is  the  an- 
swer.    Hence,  when 

the  multiplier  is  a  composite  number, 

Multiply  by  its  factors  in  succession. 

Examples. 

2.  Bought  90  hogsheads  of  sugar,  each  weighing  12cwt.  2qr. 
111b. ;  what  was  the  weight  of  the  whole  ? 

3.  What  cost  18  sheep  at  5s.  9Jd.  apiece  ? 

4.  What  cost  21  yards  of. cloth  at  9s.  lid.  per  yard? 

5.  What  cost  22  hats  at  lis.  6d.  each  ? 

6.  If  1  share  in  a  certain  stock  be  valued  at  13£.  8s.  9^-d., 
what  is  the  value  of  96  shares  ? 

7.  If  1  spoon  weighs  3oz.  5pwt.  15gr.,  what  is  the  weight  of 
120  spoons  ? 

8.  If  a  man  travel  24m.  7fur.  4rd.  in  1  day,  how  far  will  he 
go  in  1  month  ? 

9.  If  the  earth  revolve  0°  15'  per  minute,  how  far  does 
it  revolve  per  hour  ? 

10.  Multiply  39A.  3R.  17p.  30yd.  8ft.  lOOin.  by  32. 

11.  If  a  man  be  2d.  5h.  17m.  19sec.  in  walking  1  degree, 
how  long  would  it  take  him  to  walk  round  the  earth,  allowing 
3  65  J  days  to  a  year  ? 

149.  When  the  multiplier  is  not  a  composite  number,  and 
exceeds  12 ;  or  when  a  composite  number  one  of  whose  factors 
exceeds  12. 

Ex.  1.  What  is  the  value  of  453  tons  of  iron  at  18£.  17s. 
lid.  a  ton? 

N    11 


122  MULTIPLICATION   OP   COMPOUND   NUMBERS. 

OPERATION. 

£.       s.       d.  £.      s.         d. 

lton=  18    17    1  1,  X  3  =  5613       9  =  Value  of  3  tons. 

10 


10  tons  =      188    19        2,XO=      9  44    15    1  0  =  Value  of  50  tons. 

H) 

100tons=1889    11       8,X4=7558       6       8  —  Value  of  400  tons. 

Ans.    8559    16       3=  Value  of  453  tons. 

Since  453  is  not  a  composite  number,  we  cannot  resolve  it  into 
factors ;  but  we  may  separate  it  into  parts,  and  find  the  value  of  each 
part  separately:  Thus,  453  =  400  -{-  50  -|-  3.  In  the  operation,  we 
first  multiply  by  10,  and  obtain  the  value  of  10  tons,  and  this  product 
we  multiply  by  10,  and  obtain  the  value  of  100  tons.  Then,  to  find 
the  value  of  400  tons,  we  multiply  the  last  product  by  4  ;  and  to  find 
the  value  of  50  tons,  we  multiply  the  value  of  10  tons  by  5  ;  and  to 
find  the  value  of  3  tons,  we  multiply  the  value  of  1  ton  by  3.  Add- 
ing the  several  products,  we  obtain  8559£.  16s.  3d.  for  the  answer. 
Hence, 

Having  resolved  the  multiplier  into  any  convenient  parts,  as  of  units, 
tens,  §*c,  multiply  by  these  several  parts,  and  add  together  the  products 
thus  obtained  for  the  required  result. 

Examples. 

2.  Multiply  2hhd.  19gal.  Oqt.  lpt.  by  39. 

3.  Multiply  3bu.  lpk.  4qt.  lpt.  lgi.  by  53. 

4.  Multiply  16ch.  7bu.  2pk.  Oqt.  Opt.  by  17. 

5.  What  will  57  gallons  of  wine  cost  at  8s.  3£d.  per  gallon  ? 

6.  Bought  29  lots  of  wild  land,  each  containing  117 A.  3R. 
27p. ;  what  were  the  contents  of  the  whole  ? 

7.  Bought  89  pieces  of  cloth,  each  containing  37yd.  3qr. 
2na.  2in. ;  what  was  the  whole  quantity  ? 

8.  Bought  59  casks  of  wine,  each  containing  47gal.  3qt. 
lpt. ;  what  was  the  whole  quantity  ? 

9.  If  a  man  travel  17m.  3fur.  13rd.  14ft.  in  one  day,  how 
far  will  he  travel  in  a  year  ? 

10.  If  a  man  drink  3gal.  lqt.  lpt.  of  beer  in  a  week,  how 
much  will  he  drink  in  52  weeks  ? 

11.  There  are  17  sticks  of  timber,  each  containing  37ft. 
978in. ;  what  is  the  whole  quantity  ? 

12.  There  are  17  piles  of  xwood,  each  containing  7  cords  98 
cubic  feet ;  what  is  the  whole  quantity  ? 


DIVISION   OP   COMPOUND   NUMBERS.  123 


DIVISION  OF    COMPOUND  NUMBERS. 

150.  Division  of  Compound  Numbers  is  the  process  of 
dividing  compound  numbers  into  any  proposed  number  of 
equal  parts. 

Ex.  1.  Divide  139£.  13s.  lid.  2far.  equally  between  5  per- 
sons. Ans.  27£.  18s.  9d.  2far. 

operation.  Having  divided  139£.  by  5,  we  find 

£.        s.        d.      far.     the  quotient  to  be  27£.,  and  4£.  re- 

5)139     13     11     2      maining.   We  place  the  quotient  27£. 

— under  the  139£.,  and  the  remainder 

2  7  18  9  2  4£.  reduced  to  shillings  =  80s. ;  80s. 
-f-  the  13s.  in  the  dividend  =  93s. ; 
93s.  -5-  5  =  18s.  and  a  remainder  of  3s.  We  write  the  quotient 
18s.  under  the  shillings  in  the  dividend;  and  the  remainder  3s.  re- 
duced to  pence  =  36d. ;  36d.  -f-  lid.  in  the  dividend  =  4 7d. ; 
47d.  -5-  5  —  9d.  and  a  remainder  of  2d.  We  write  the  quotient  9d. 
under  the  pence  in  the  dividend ;  and  the  remainder  2d.  reduced  to 
farthings  =  8far.,  -\-  the  2far.  in  the  dividend  =  lOfar. ;  lOfar.  -h  5 
=  2far.  The  quotient  2far.  we  write  under  the  farthings  in  the 
dividend;  and  thus  find  the  answer  to  be  27£.  18s.  9d.  2far. 

Rule.  — Divide  as  in  division  of  simple  numbers,  each  denomination 
in  its  order,  beginning  with  the  highest. 

If  there  be  a  remainder,  reduce  it  to  the  next  lower  denomination, 
adding  in  the  number  already  contained  in  the  dividend  of  this  de- 
nomination, if  any,  and  divide  as  before. 


Proof.  —  The  same  as  in  simple  numbers. 

Note.  —  When  the  divisor  and  dividend  are  both  compound  numbers,  they 
must  be  reduced  to  the  same  denomination,  and  the  division  then  is  that  of 
simple  numbers. 

Examples. 

Note.  —  The  answers  to  the  following  examples  are  found  in  the  corre- 
sponding numbers  of  examples  in  Multiplication  of  Compound  Numbers. 


5) 


2. 

£.          s.        d. 

8)19     10    4 

T.          cwt. 

9)557     19 

3. 

qr.      lb.          oz. 

11115 

dr. 
7 

4. 

lb.         oz.        pwt.     gr. 

)  39     10     13     3 

lb. 

8)261 

5. 

oz.      pwt.    gr. 

110     0 

9. 

A. 

R.        p. 

yd.    ft. 

in. 

181 

3     11 

6      4 

41 

124  DIVISION   OP  COMPOUND  NUMBERS. 

6.  .  7. 

ft)  <5         5      3       gr.  deg*  m-      fur-    rd«    ft*     in- 

11)427  10  0  2  14   12)858  44  4  6  7  0 

8. 

m.    fur.  ch.  p.   1. 

12)215  7  9  3  1    1 

10.  Divide  54yd.  2qr.  3na.  equally  among  5  persons. 

11.  Divide  123tun  3hhd.  36gal.  3qt.  by  7. 

12.  Divide  209hhd.  55gal.  3qt.  Opt.  lgi.  by  7. 

13.  What  is  the  value  of  118bu.  lpk.  5qt.  -r-  6  ? 

14.  What  is  the  value  of  HOy.  343d.  3h.  41m.  12s.  —  8  ? 

15.  Divide  149deg.  9m.  5fur.  13rd.  3yd.  1ft.  by  9. 

16.  A  man  divides  his  farm  of  214A.  3R.  12p.  equally 
among  his  9  sons ;  how  much  does  each  receive  ? 

17.  If  one  man  perform  a  certain  piece  of  labor  in  3d.  16h. 
54m.,  how  long  would  it  take  12  men  to  perform  the  same 
work  ? 

18.  A  farmer  has  29  bushels  of  oats  which  he  wishes  to 
put  in  8  sacks  ;  how  much  must  each  sack  contain  ? 

151 ,  When  the  divisor  is  a  composite  number,  and  none  of 
its  factors  exceed  12. 

Ex.  1.  If  35  loads  of  coal  weigh  72T.  14cwt.  2qr.  101b., 
what  will  1  load  weigh  ? 

operation.  We  find  the  fac- 

T.      cwt.    qr.     lb.  tors  0f  35  to  be  5 

5)72     14     2     10  =  weight  of  35  loads,     and  7.    We  there- 

7  )  14     10     3     17=  weight  of  7  loads.     ^  **  ^ 

2         12         6  =  weight  of  1  load.       by   5,   and    obtain 

the  weight  of  7 
loads ;  and  the  weight  of  7  loads  we  divide  by  7,  and  thus  find  the 
weight  of  1  load.     Hence,  when  the  divisor  is  a  composite  number, 

Divide  hy  its  factors  in  succession. 

Examples. 

2.  If  90  hogsheads  of  sugar  weigh  56T.  14cwt.  3qr.  151b., 
what  is  the  weight  of  1  hogshead  ? 

3.  What  will  be  the  price  of  1  sheep,  if  18  cost  5£.  4s.  3d.  ? 


DIVISION   OF   COMPOUND   NUMBERS.  125 

4.  If  21  yards  of  cloth  cost  10£.  8s.  3d.,  what  is  the  price 
of  1  yard  ? 

o.  What  is  the  value  of  1  hat,  when  22  cost  12£.  13s.  Od.  ? 

6.  When  96  shares  of  a  certain  stock  are  valued  at  1290£. 
4s.  Od.,  what  would  be  the  cost  of  1  share  ? 

7.  If  120  spoons  weigh  321b.  9oz.  15pwt.,  what  does  1 
weigh  ? 

8.  If  a  man  in  1  month  travels  746m.  5fur.,  how  far  does 
he  go  in  1  day  ? 

9.  If  the  earth  revolves  15°  on  its  axis  in  1  hour,  how  far 
does  it  revolve  in  1  minute  ? 

10.  Divide  1275A.  2E.  16p.  22yd.  8ft.  32in.  equally  among 
32  men. 

11.  If  a  man  walk  round  the  earth  in  2y.  68d.  19h.  54m., 
how  long  would  it  take  him  to  walk  1  degree,  allowing  3  65  J 
days  to  a  year  ? 

152»  When  the  divisor  is  not  a  composite  number,  and 
exceeds  12,  or  when  a  composite  number  one  of  whose  factors 
exceeds  12,  the  whole  operation  can  he  written,  as  in  the  follow- 
ing example. 

Ex.  1.  Divide  360£.  8s.  4d.  by  173.        Ans.  2£.  Is.  8d. 

OPERATION. 
£.  S.         d. 

17  3)360     8     4  (  2£.         We  divide  the  pounds  by  173,  and 

3  4  6  obtain  2£.  for  the  quotient,  and  14£. 

remaining,  which  we  reduce  to  shil- 

1  4  lings,  and  add  the  8s.,  and  again  divide 

2  0  by  173,  and  obtain  Is.  for  the  quotient. 

i  7  o  \  9  o  o  /  i  The  remainder,  115d.,  we  reduce  to 

U^J^««(is.  pence,  and  add  the  4d.,  and  again  di- 

1  *  3  vide  by  173,  and  obtain  8d.  for  the 

\\§  quotient.     Thus,   the  method  is    the 

^  2  same  as  by  general  rule  (Art.  150). 

By  writing  the  several  quotients,  we 

173)1384(8d.  obtain  2£.  Is.  8d.  for  the  answer. 
1384 

2.  Divide  89hhd.  52gal.  3qt.  lpt.  by  39. 

3.  Divide  179bu.  3pk.  5qt.  Opt.  lgi.  by  53. 

4.  Divide  275ch.  19bu.  2pk.  equally  among  17  men. 

11* 


126  PRINCIPLES   AND   APPLICATIONS. 

5.  If  57  gallons  of  wine  cost  23£.  lis.  5£d.,  what  cost  1 
gallon  ? 

6.  Divide  3419A.  2R.  23p.  by  29. 

7.  If  89  pieces  of  cloth  contain  3375yd.  3qr.  lna.  0£in., 
how  much  does  1  piece  contain  ? 

8.  If  59  casks  contain  44hhd.  52gal.  2qt.  lpt.  of  wine,  what 
are  the  contents  of  1  cask  ? 

9.  If  a  man  travel  in  1  year  (365  days)  6357m.  5fur.  14rd. 
11^-ft.,  how  far  is  that  per  day  ? 

10.  When  175gal.  2qt.  of  beer  are  drunk  in  52  weeks,  how 
much  is  consumed  in  1  week  ? 

11.  When  17  sticks  of  timber  measure  15T.  38ft.  1074in., 
how  many  feet  does  1  contain  ? 

12.  Divide  132  cords  2ft.  by  17. 

13.  Divide  697T.  18cwt.  3qr.  141b.  by  146. 

Ans.  4T.  15cwt.  2qr.  10-J-fflb. 

14.  Divide  916m.  3fur.  30rd.  10ft.  6in.  by  47. 

Ans.  19m.  3fur.  39rd.  13ft.  2|fin. 

15.  Divide  718A.  3R.  37p.  by  29.     Ans.  24A.  3R.  6Jfp. 

16.  Divide  815A.  1R.  17p.  200ft.  by  87. 

Ans.  9A.  1R.  19p.  139|fft. 

17.  Divide  144A.  3R.  18p.  3yd.  1ft.  36in.  by  11.  4 

Ans.  13A.  OR.  27p.  3yd.  Oft.  45T9Tin. 


PRINCIPLES  AND  APPLICATIONS. 

DIFFERENCE  BETWEEN  DATES. 

153.     To  find  the  time  between  two  different  dates. 
Ex.  1.     What  is  the  difference  of  time  between  May  16, 
1819,  and  March  4,  1857  ?  Ans.  37y.  9mo.  18d. 

Commencing    with    January, 

first  operation.  the  first  month  in  the  year,  and 

y-         m0*      d-  counting  the  months  and  days  in 

Min.      18  5  7     2         4         the  later  date  up  to  March  4th, 

Sub.      1819      4      16  we  find  that  2mo.  and  4d.  have 

elapsed.    We  therefore  write  the 

Rem.  3  7     9      18  numbers  for  subtraction  as  in  the 

first  operation. 


SECOND   OPERATION. 

Min.     18  5  7     3 
Sub.     1819     5 

4 
16 

Rem.          3  7     9 

18 

PRINCIPLES    AND    APPLICATIONS.  127 

The    same    result,    however, 
could  be  obtained,  as  some  pre- 
fer,   by   reckoning   the   number 
of  the  given  months  instead  of 
the  number  of  months  that  have 
elapsed  since  the  beginning  of 
the  year,  which  would  require  the  numbers  to  be  written  as  in  the 
second  operation.     Written  either  way,  the  earlier  date,  being  placed 
under  the  later,  is  subtracted  from  it. 

Note.  —  In  finding  the  difference  between  two  dates,  and  in  computing  in- 
terest for  less  than  a  month,  30  days  are  considered  a  month.  In  legal  trans- 
actions, however,  a  month  is  reckoned  from  any  day  in  one  month  to  the  same 
day  of  the  following  month.  The  above  process,  which  is  that  ordinarily  used 
by  business  men,  does  not  give  always  the  exact  time  between  two  different 
dates.  The  result  obtained  by  it  may  deviate  sometimes  a  day,  and,  less  often, 
two  days,  from  the  exact  difference.  But  for  practical  purposes  it  is  generally 
regarded  as  sufficiently  accurate. 

Examples. 

2.  What  is  the  time  from  June  3d,  1854,  to  April  19th,  1857? 

3.  A  note  was  given  October  26th,  1856,  and  paid  June  12th, 
1857 ;  how  long  was  it  on  interest  ? 

4.  The  Pilgrims  landed  at  Plymouth  December  22d,  1620, 
N.  S.,  and  the  Declaration  of  Independence  was  made  July  4th, 
1776  ;  what  is  the  difference  of  time  between  these  events? 

5.  General  Washington  was  born  February  22d,  1732,  and 
died  December  14th,  1799 ;  how  long  did  he  live? 

Ans.  67y.  9mo.  22d. 
154.     To  find  the  exact  number  of  days  between  two  differ- 
ent dates. 

Ex.  1.  How  many  days  from  January  28  to  July  30,  com- 
mon year?  Ans.  183  days. 

OPERATION. 

January  to  July  =  6mo.  6  X  31  =186  days. 

For  Feb.  3d.,  April  Id.,  June  Id.,     3  +  1  +  1=         5     " 

Tsi 

For  difference  between  30  and  28,         30  —  28  =        2 

Ans.  18  3  days. 

The  difference  between  January  and  July  we  find  to  be  6 mo., 
which,  multiplied  by  31,  the  greatest  number  of  days  in  any  month  in 
the  year,  gives  186  days.  But  since  in  the  interval  of  time  in- 
cluded between  the  given  dates  several  months  end  that  do  not  con- 


128  PRINCIPLES    AND    APPLICATIONS. 

tain  31  days,  we  make  deductions  for  these,  which,  in  all,  amount  to 
5  days,  and  have  left  181  days ;  and  as  the  difference  between  the 
given  dates  is  the  difference  between  30  and  28  more  than  exactly 
6mo.,  we  add  2  to  the  281  days,  thus  obtaining  183  days,  the  differ- 
ence of  time  required.  Hence,  to  find  the  number  of  days  between 
two  different  dates, 

Find  the  number  of  months  ending  between  the  given  dates,  and  mul- 
tiply that  number  by  31,  and  from  the  product  make  the  necessary  de- 
duction for  the  months  counted  that  do  not  contain  31  days,  if  any. 
Should  the  later  date  end  later  in  the  month  than  the  earlier,  add  the 
difference  of  days  ;  but  should  it  end  earlier,  subtract  the  same. 

Note.  —  The  exact  difference  in  days  between  two  different  dates  can  also 
be  obtained  by  use  of  the  table  in  Note  2,  Art.  142. 

Examples. 

2.  How  many  days  has  a  note  to  run  dated  November  15, 
1856,  and  made  payable  February  13, 1857  ?    Ans.  90  days. 

3.  How  many  days  from  June  18,  1855,  to  May  1,  1856  ? 

4.  How  many  days  from  March  4  to  May  3  of  the  same 
year? 

5.  From  November  4,  1856,  to  April  4,  1857,  how  many 
days?  Ans.  151  days. 

6.  In  a  leap  year,  how  many  days  are  there  from  the  7th  of 
January  to  the  11th  of  December  ?  Ans.  339  tlays. 

155.  To  find  the  day  of  the  week  corresponding  to  any  given 
day  of  the  month,  when  the  day  of  the  week  of  some  other  date 
is  given. 

Ex.  1.  If  the  16th  day  of  May  be  on  Saturday,  what  day 
of  the  week  will  the  next  25th  of  December  be  ? 

Ans.  Friday. 


OPERATION. 


From  May  16  to  December  25  =  223  days. 
223  days  -r-  7  =  31  weeks  and  6  days. 
6  days  after  Saturday  =  Friday,  Ans. 

Having  found  the  difference  of  time  in  days  between  the  given 
dates,  we  bring  the  days  to  weeks  by  dividing  by  7,  and  obtain  31 
weeks  and  6  days.  The  25th  of  December,  therefore,  must  come  6 
days  after  Saturday,  or  on  Friday.     Hence,  we 

Reduce  the  days  between  the  given  dates  to  weeks.  Should  there  be 
no  remainder,  the  day  given  will  be  the  same  as  that  sought,  but  should 
there  be  a  remainder,  it  will  indicate  the  number  of  days  that  the  day 
sought  is  after  the  day  given. 


PRINCIPLES    AND    APPLICATIONS.  129 

Examples. 

2.  If  the  2d  day  of  April  be  on  Wednesday,  what  day  of 
the  week  will  the  following  4th  of  July  be  ?        Ans.  Friday. 

3.  If  a  leap  year  commence  on  Tuesday,  on  what  day  will 
the  17th  of  June,  the  anniversary  of  the  battle  of  Bunker  Hill, 
happen  ? 

4.  If  in  a  common  year  the  25th  of  December,  or  Christmas, 
be  on  Tuesday,  on  what  day  did  the  year  commence  ? 

Ans.  Monday. 

5.  If  the  4th  day  of  November  be  on  Tuesday,  what  day 
of  the  next  February  will  be  the  second  Monday  of  that 
month?  Ans.    The  9th. 

6.  A  bill  was  dated  on  Thursday,  December  20th,  1855, 
and  made  payable  90  days  after  date.  In  what  year  and 
month,  and  on  what  day  of  the  month  and  week,  did  it  become 
due?  Ans.  Wednesday,  March  19,  1856. 

DEFFEKENCE  OF  LATITUDE. 

156.  Latitude  is  the  distance  of  any  place  from  the  equator, 
north  or  south.  It  is  reckoned  in  degrees,  minutes,  and  sec- 
onds, from  the  equator  to  either  pole  of  the  earth ;  and  cannot 
exceed  90  degrees,  or  one  fourth  of  the  earth's  circumference. 
Places  north  of  the  equator  are  said  to  be  in  north  latitude, 
and  those  south  of  the  equator,  in  south  latitude. 

Note.  —  The  shape  of  the  earth  not  being  that  of  a  perfect  sphere,  but  some- 
what flattened  toward  the  poles,  the  degrees  of  latitude  differ  a  little  from  each 
other  in  length  toward  either  pole.  Thus  the  1st  degree  is  about  68^  statute 
miles  in  length;  the  40th  degree,  about  68jq8q  miles;  and  the  89th  degree  about 
69^  miles. 

157*     To  find  the  difference  of  latitude  of  any  two  places. 

Ex.  1.  The  latitude  of  London  is  51°  SV  north,  and  that 
of  Boston  is  42°  23'  north.  What  is  their  difference  of  lati- 
tude ?  Ans.  9°  8'. 

operation.  The    two    places    being 

Lat.  of  London  =  5  1°     3  V  N.     both  in  north  latitude,  we 
Lat.  of  Boston   =4  2°     2  3'  N.     ^tract    the    less   latitude 

from  the  greater,     it  the 

Dif.  of  Latitude  =     9°        8'  N.     latitude  of   the  one  place 

had  been  north  and  that  of 


130  PRINCIPLES    AND    APPLICATIONS. 

the  other  south,  we  should  have  added  the  two  latitudes  together  for 
the  difference  required.     Hence 

When  the  latitudes  of  two  places  are  either  both  north  or  both  south, 
subtract  the  less  latitude  from  the  greater  for  their  difference;  and 
when  the  latitude  of  the  one  place  is  north  and  the  other  south,  add  the 
latitudes  together  for  their  difference. 

Note.  —  In  north  latitude,  when  the  sailing  is  southerly,  or  in  south  latitude, 
when  the  sailing  is  northerly,  if  the  difference  of  latitude  be  subtracted  from 
the  latitude  left,  the  remainder  will  be  the  latitude  in  ;  or,  in  north  latitude, 
when  the  sailing  is  northerly,  or  in  south  latitude,  when  the  sailing  is  southerly, 
if  the  difference  of  latitude  be  added  to  the  latitude  left,  the  sum  will  give 
the  latitude  in. 

Examples. 

2.  The  latitude  of  Quebec  is  46°  48'  north,  and  that  of  New 
Orleans  29°  57'  north.  What  is  their  difference  of  lati- 
tude ?  Ans.  16°  51'. 

3.  The  latitude  of  Washington  City  is  38°  53' north,  and  that 
of  Cape  Horn  55°  58'  south.  What  is  their  difference  of  lati- 
tude ?  Ans.  94°  51'. 

4.  Valparaiso  is  in  latitude  33°  2'  south,  and  San  Francisco 
37°  48'  north.     What  is  their  difference  of  latitude  ? 

5.  Captain  James  Francis,  sailing  southerly  from  New  York 
City,  whose  latitude  is  40°  42'  north,  found  on  reaching  Havana 
that  his  latitude  differed  17°  33'  from  that  he  left.  What  lati- 
tude was  he  then  in  ?  Ans.  23°  9'  north. 

6.  Philadelphia  is  9°  15'  of  latitude  north  of  Mobile, 
whose  latitude  is  30°  41'  north ;  what  is  the  latitude  of  Phila- 
delphia ?  Ans.  39°  56'  north. 

DIFFERENCE  OF  LONGITUDE. 

158.     Longitude  is  the  distance  of  any  place  from  a  given 

meridian,  east  or  west.     It  is  reckoned  in  degrees,  minutes, 

and  seconds,  and  cannot  exceed  180  degrees,  or  one  half  of  a 

circle. 

Note  1.  —  A  degree  of  longitude  on  the  equator  is  about  69^  statute  miles, 
and,  in  general,  a  degree  is  5JC  of  any  circle  of  latitude.  The  meridians  all  con- 
verge from  the  equator  to  the  poles  to  a  point,  so  that  the  degrees  of  longi- 
tude under  different  parallels  of  latitude  vary,  diminishing  with  the  circles  of 
latitude,  till  at  the  poles  the  longitude  becomes  nothing.    (Art.  133,  Note  4.) 


PRINCIPLES    AND    APPLICATIONS.  131 

Note  2.  —  The  mariners  of  Great  Britain  and  the  United  States  reckon 
longitude  from  the  meridian  of  Greenwich  in  England,  as  do  the  nautical  books 
of  both  countries  for  the  most  part.  American  maps,  however,  very  generally 
have  longitude  reckoned  both  from  the  meridian  of  Greenwich,  and  from  the 
meridian  of  Washington. 

159.     To  find  the  difference  of  longitude  of  any  two  places. 
Ex.  1.    What  is  the  difference  of  longitude  between  Boston, 
which  is  71°  4'  west,  and  Detroit,  which  is  82°  58'  west. 

Ans.  11°  54'. 
operation.  The    two    places    being 

Long,  of  Detroit    =    8  2°     5  8'      both  in  west  longitude,  we 
Long,  of  Boston     =7  1°        4'      subtract  the  less  longitude 

°  from  the  greater.     If,  how- 

Dif.  of  Longitude  =11°     $4?      ever,  one  of  the  places  had 

been  in  east  longitude,  and 
the  other  in  west,  we  should  have  added  the  two  longitudes  together 
for  the  difference  required.    Hence, 

When  the  longitudes  of  two  places  are  either  both  east  or  both  west,  sub- 
tract the  less  longitude  from  the  greater  for  their  difference ;  and  when 
the  longitude  of  the  one  place  is  east,  and  that  of  the  other  west,  add  the 
longitudes  together  for  their  difference. 

Note.  —  In  adding  together  two  longitudes,  should  their  sum  exceed  180 
degrees,  subtract  it  from  360  degrees,  and  the  remainder  will  be  the  correct 
difference  of  longitude. 

Examples. 

2.  What  is  the  difference  of  longitude  between  the  city  of 
Washington,  whose  longitude  is  77°  16-'  west,  and  Paris,  whose 
longitude  is  2°  20'  east?  Ans.  79°  36'. 

3.  The  United  States  extend  from  the  St.  Croix  River, 
longitude  67°  2'  west,  to  Cape  Flattery,  longitude  124°  43' 
west.  Over  how  many  degrees  of  longitude  does  the  Union 
extend  ? 

4.  What  is  the  difference  of  longitude  between  Raleigh, 
whose  longitude  is  78°  48'  west,  and  Sacramento  City,  whose 
longitude  is  120°  west  ?  Ans.  41°  12'. 

5.  What  is  the  difference  of  longitude  between  Hartford, 
whose  longitude  is  72°  40'  west,  and  Fort  Leavenworth  whose 
longitude  is  94°  44'  west  ? 

6.  The  longitude  of  Honolulu  is  157°  52'  west,  and  that 
of  Canton  113°  14'  east.  What  is  their  difference  of  longi- 
tude ?  Ans.  88°  54'. 


132  PRINCIPLES  AND   APPLICATIONS. 

LONGITUDE  AM)   TIME. 

160»  To  find  the  time  corresponding  to  degrees  and  min- 
utes of  longitude. 

Ex.  1.   The   difference  of  longitude  between   Boston  and 

London  being  71°  4',  what  is  their  difference  of  time  ? 

Ans.  4h.  44m.  16sec. 

operation.  Sincc  |«  0f  longitude 

Dif.  of  longitude  =      71°     4'  corresponds  to  4sec.  of 

4  time,  and  1°  of  longitude 

tw.    r  ±-  ^       7*  to  4m.  of  time  (Art.  143, 

Dif.  of  time  =  284m.  16sec.  Note  2)>  4,  covrresponds 

284m.  16sec.  =  4h.  44m.  16sec.  Ans.     to    16sec,  and    71°   to 

284m.;  and  284m.  16sec. 
=  4h.  44m.  16sec.,  the  answer  required.  The  apparent  motion 
of  the  sun  being  west,  the  time  at  Boston  is  as  much  earlier  than 
that  of  London  as  the  difference  of  time  between  them.  Thus,  when 
it  is  noon  at  London,  it  wants  4h.  44m.  16sec.  of  noon  at  Boston 
(Art.  143,  Note  2).  Therefore,  to  find  the  time  corresponding  to 
degrees  and  minutes  of  longitude, 

Multiply  the  minutes  of  longitude  by  4,  for  seconds  of  time. 
Multiply  the  degrees  of  longitude  by  4,  for  minutes  of  time. 

Note  1.  —  Should  the  seconds  of  time,  when  found,  be  60  or  more,  they  may- 
be reduced  to  minutes;  and  should,  also,  the  minutes  be  60  or  more,  they  may 
be  reduced  to  hours.  The  difference  of  time  between  two  places  is  exactly  as 
much  as  the  true  clock  time  of  the  one  is  fast  or  slow  as  compared  with  that 
of  the  other. 

Examples. 

2.  Galveston  in  Texas  is  14°  43'  west  of  Pittsburg.  When 
it  is  12  o'clock  at  Galveston,  what  is  the  time  at  Pittsburg  ? 

3.  The  longitude  of  Valparaiso  is  71°  37'  west,  and  the  lorn 
gitude  of  Rome  is  20°  30'  east.  When  it  is  llh.  15m.  A.  M, 
at  Valparaiso,  what  is  the  time  at  Rome  ? 

Ans.  23m.  28sec.  past  5  o'clock,  P.  M. 

4.  The  longitude  of  Jerusalem  is  35°  32'  east,  and  the  Ion, 
gitude  of  Baltimore  76°  37'  west.  When  it  is  9  o'clock,  A.M, 
at  Jerusalem,  what  time  is  it  at  Baltimore  ? 

Ans.  lh.  31m.  24sec.  A.  M. 

161  •  To  find  the  longitude  corresponding  to  hours,  minutes^ 
and  seconds  of  time. 


PRINCIPLES   AND   APPLICATIONS.  133 

Ex.  1.  The  difference  of  time  between  Boston  and  London 
is  4h.  44m.  16sec. ;  what  is  the  difference  of  longitude  ? 

Ans.  71°  4'. 

operation.  Since  lh.  of  time  corresponds  to 

1  5°  X  4  =  6  0°  15o  of  longitude,  4m.  of  time  to  1° 

4  4-7-4=11°  of  longitude,  and  4sec.  of  time  to 

1  6    -j-  4  ==  4/      1'  of  longitude  (Art.  143,  Note  2), 

4h.  corresponds  to  60°,  44m.  to  11°, 

Dif.  of  longitude  =7  1°  4'      and  16sec.  to  4';  and  60°+  ll°-f 

4/  =  71°  4',  the  answer  required. 
Hence,  to  find  the  longitude  corresponding  to  hours,  minutes,  and 
seconds  of  time, 

Multiply  the  hours  of  time  by  15,  and  divide  the  minutes  of  time  by  4, 
for  degrees  of  longitude;  and  divide  the  seconds  of  time  by  ±,for  min- 
utes of  longitude.  The  several  results  added  together  will  be  the  differ- 
ence of  longitude. 

Examples. 

2.  The  difference  of  time  between  Washington  and  Cincin- 
nati is  29m.  36sec. ;  what  is  the  difference  of  longitude  ? 

3.  A  ship-captain  sailing  from  New  York  to  Europe,  after 
being  at  sea  some  days,  on  taking  an  observation,  found  that 
the  sun  at  noon  was  2h.  20m.  40sec.  earlier  than  the  New 
York  time,  as  shown  by  his  chronometer.  How  many  degrees 
east  had  he  sailed  ?  Ans.  35°  10'. 

4.  A  gentleman  travelling  west  from  Philadelphia,  whose 
longitude  is  75°  10'  west,  found,  on  arriving  at  St.  Louis,  that 
his  watch,  an  accurate  timekeeper,  which  was  right  when  he  left 
Philadelphia,  was  lh.  20sec.  earlier  than  the  time  at  St.  Louis. 
What,  then,  is  the  longitude  of  St.  Louis  ?      Ans.  90°  15/  west. 

5.  The  difference  of  time  between  Baltimore  and  New  Or- 
leans is  53m.  30sec. ;  what  is  the  difference  of  longitude  ? 

6.  When  it  is  noon  at  St.  Paul's,  longitude  93°  b'  west,  it  is 
at  Bangor  lh.  37m.  12sec.  P.M.;   what  is  the  longitude  of 

» Bangor  ?  Ans.  68°  47'  west. 

7.  When  it  is  11  A.  M.  at  a  place  30°  east  of  Greenwich, 
it  is  3h.  44m.  20sec.  A.  M.  at  Buffalo ;  what  is  the  longitude 
of  Buffalo  ?  Ans.  78°  55'  west. 

8.  The  difference  of  time  between  Cambridge  Observatory 
and  Greenwich  is  4h.  44m.  32sec. ;  what  is  the  difference  of 
longitude?  Ans.  71°  8'. 

N    12 


134  MISCELLANEOUS   EXAMPLES. 


MISCELLANEOUS   EXAMPLES. 

1.  If  the  population  of  the  world  be  as  follows:  America, 
57,650,000;  Europe,  263,517,496;  Asia,  626,400,000;  Africa, 
100,000,000 ;  Australia,  1,445,000  ;  Polynesia,  1,500,000 ;  and 
the  average  length  of  life  be  33  years,  what  must  be  the  aver- 
age number  of  deaths  annually  ?  Ans.  31,833,712. 

2.  A  farmer  has  in  3  bins  755  bushels  of  grain;  there 
being  in  the  first  125  bushels,  and  in  the  second  96  bushels, 
more  than  in  the  third ;  how  many  bushels  in  the  second  and 
third  ?  Ans.  363  in  the  second,  267  in  the  third. 

3.  There  is  a  certain  island  30  miles  in  circumference.  If 
A  and  B  commence  travelling  round  it,  A  at  the  rate  of  3 
miles  an  hour,  and  B  at  the  rate  of  5  miles  an  hour,  how  far 
apart  will  they  be  at  the  end  of  30  hours  ? 

4.  Having  money  to  invest,  I  purchased  two  farms  at  $  1,750 
each,  and  19  shares  of  bank  stock  at  $  103  per  share,  and  have 
left  $  113  ;  how  much  money  had  I  ?  Ans.  $  5,570. 

5.  It  has  been  agreed  by  12  men  to  gather  960  bushels  of 
cranberries,  and  receive  for  their  labor  one  half  of  the  quantity 
gathered ;  after  one  half  was  gathered,  one  third  of  the  men 
withdrew,  leaving  the  others  to  complete  the  job.  How  many 
bushels  should  each  man  receive  ?  Ans.  Those  who  left,  20 
bushels  each ;  those  who  remained,  50  bushels  each. 

6.  A  drover  made  $  652.00  by  selling  a  lot  of  sheep,  at  a 
profit  of  50  cents  each  ;  how  many  did  he  sell  ? 

7.  What  will  it  cost  to  carpet  a  floor  that  is  18  feet  wide 
and  27  feet  long,  provided  the  carpeting  cost  $  2.25  per  yard  ? 

8.  If  a  young  man,  by  early  rising  and  economy  of  time, 
can  save  for  study  and  improvement  of  mind  two  and  a  half 
hours  a  day,  how  many  years'  study,  of  12  hours  per  day,  can 
thus  be  gained  in  20  years  ?  Ans.  4y.  60d.  10^-h. 

9.  From  4  piles  of  wood,  the  first  containing  7c.  76ft.  1671in., 
the  second  16c.  28ft.  56in.,  the  third  29c.  127ft.  1000in.,  the 
fourth  29c.  10ft.  121 6in.,  I  have  sold  45  cords  and  6  cord  feet; 
how  much  remains  ?  #  Ans.  37c.  19ft.  487in. 

10.  Boston  is  in  north  latitude  42°  21'  ;  Portland  is  in  lati- 
tude  1°  15'  north  of  Boston ;  and  Charleston  is  in  latitude 


MISCELLANEOUS   EXAMPLES.  135 

10°  40'  south  of  Portland.     What  is  the  latitude  of  Charles- 
ton? Ans.  32°  5 6'  north. 

11.  If  1  cubic  foot  of  anthracite  coal  weighs  54  pounds, 
how  many  cubic  feet  of  space  are  required  to  stow  2  tons  of 
2000  pounds  each  ?  Ans.  74/T  cubic  feet. 

12.  Two  engineers,  A  and  B,  surveyed  a  certain  house-lot. 
A  made  its  contents  3R.  18p.  0yd.  6ft.  64sq.  in.,  but  B  made 
its  contents  3R.  17p.  30yd.  8ft.  lOOsq.  in.  How  much  did  the 
one  differ  from  the  other  ? 

13.  The  products  of  the  industry  of  250,000  persons  in  Mas- 
sachusetts, during  the  year  1855,  amounted  to  $  295,300,000. 
What  was  the  average  amount  to  each  individual,  and  how 
much  was  added  to  the  capital  of  the  State,  if  one  fourth  of 
the  whole  amount  was  saved  ?  Ans.  $  1,181.20  to  each  in- 
dividual ;  $  73,825,000  added  to  the  capital. 

14.  The  capacity  of  a  certain  cistern  is  216  cubic  feet ;  how 
many  hogsheads  of  water  will  it  contain  ? 

15.  What  day  of  the  month  and  what  day  of  the  year  is  the 
second  Monday  of  May,  in  a  common  year  commencing  on 
Thursday  ?      Ans.  11th  day  of  May  ;  131st  day  of  the  year. 

16.  Purchased  18T.  17cwt.  3qr.  201b.  of  copperas,  at  4  cents 
per  pound.  I  sold  4T.  6cwt.  lqr.  141b.  at  5  cents  per  pound, 
and  7T.  lcwt.  3qr.  101b.  at  6  cents  per  pound.  Moses  Atwood 
purchased  one  fourth  of  the  remainder  at  6  cents  per  pound. 
One  half  of  what  then  remained  I  sold  to  J.  Gale  at  10  cents 
per  pound.  The  remainder  I  sold  to  J.  Smith  at  12  cents  per 
pound ;  but  he  has  become  a  bankrupt,  and  I  lose  half  my  debt. 
What  have  I  gained  by  my  purchase  ?  Ans.  $  894.07J-. 

17.  The  distance  between  Boston  and  San  Francisco  is  2691 
miles.  If  Nathan  Swift  of  San  Francisco  and  Oliver  Fleet  of 
Boston,  on  Thursday,  the  first  day  of  January,  1857,  set  out  to 
meet  each  other,  Swift  travelling  3  miles  7  furlongs  29  rods 
15  feet  per  hour,  and  Fleet  5  miles  10  rods  and  1£  feet  per 
hour,  both  travelling  6  J  hours  per  day,  commencing  at  8  o'clock, 
A.  M.,  provided  they  rest  on  the  Sabbath,  in  what  year  and 

tonth,  and  on  what  day  of  the  month,  and  at  what  time  of  the 
day,  will  they  meet,  and  how  far  will  each  have  travelled  ? 

Ans.  On  Monday,  February  23, 1857,  2h.  30m.  P.  M.  Swift, 
1186m.  4fur.  22rd.  13ft.  6in.;  Fleet,  1504m.  3fur.  17rd.  3ft. 


136  PEINCIPLES  AND   APPLICATIONS. 

EXAMPLES  BY  ANALYSIS. 

1.  If  7  pairs  of  shoes  cost  $  8.75,  what  will  one  pair  cost? 
what  will  20  pairs  cost?  Ans.  $  25.00. 

2.  If  5  tons  of  hay  cost  $  85,  what  will  1  ton  cost  ?  what  will 
17  tons  cost?  Ans.  $289.00. 

3.  When  $  0.75  are  paid  for  3gal.  of  molasses,  what  is  the 
value  of  lgal.  ?     What  cost  37  gal.  ? 

4.  Gave  $  1.92  for  41b.  of  tea;  what  cost  lib.?  what  cost 
371b.?  Ans.  $17.76. 

5.  For  121b.  of  rice  I  paid  $  1.08  ;  what  was  paid  for  lib. ; 
and  what  must  I  give  for  251b.  ?  Ans.  $  2.25. 

6.  Gave  S.  Smith  $  63.00  for  9  tubs  of  butter ;  what  was 
the  cost  of  1  tub  ?     What  cost  27  tubs  ?  Ans.  $  189.00. 

7.  T.  Swan  can  walk  20  miles  in  5  hours  ;  how  far  can  he 
walk  in  1  hour  ?  How  long  would  it  take  him  to  walk  from 
Bradford  to  Boston,  the  distance  being  in  a  straight  line  28 
miles  ? 

8.  If  a  hungry  boy  would  eat  49  crackers  in  1  week,  how 
many  would  he  eat  in  1  day  ?  how  many  would  be  sufficient  to 
last  him  19  days?  Ans.  133  crackers. 

9.  Gave  $  20  for  5  barrels  of  flour ;  what  cost  1  barrel  ? 
what  cost  40  barrels  ?  Ans.  $  160.00. 

10.  For  31b.  of  lard  there  were  paid  36  cents ;  what  was  the 
cost  of  371b.  ? 

11.  Paid  F.  Johnson  72  cents  for  9  nutmegs  ;  how  many  cents 
were  paid  for  1  nutmeg ;  and  what  should  be  charged  for  37 
nutmegs?  Ans.  $2.96. 

12.  Paid  2£.  17s.  5d.  for  521b.  of  sugar;  what  cost  lib.? 
what  cost  761b.  ? 

13.  Paid  4£.  3s.  lid.  for  761b.  of  sugar;  what  cost  521b.? 

14.  If  a  man  walk  17m.  4fur.  28rd.  in  6  days,  how  far  will 
he  walk  in  100  days  ?  Ans.  293m.  lfur. 

15.  If  a  farmer  feed  to  his  stock  in  7  months  41  bu.  3pk.  4qt. 
lpt.  of  grain,  how  much  is  required  for  1  month  ?  how  much 
for  7  years?  Ans.  502bu.  2pk.  6qt. 

16.  A  field  containing  39 A.  2R.  5p.  8yd.  6ft.  108in.  will 
pasture  8  cows  during  the  season.  How  large  a  field  will  pas- 
ture 1  cow  ?     How  large  a  field  72  cows  ? 


PROPERTIES    OF    NUMBERS.  137 

17.  If  4  casks  of  vinegar  contain  63gal.  3qt.,  what  are 
the  contents  of  one  cask?  What  are  the  contents  of  37 
casks  ?  Ans.  589gal.  2qt.  lpt.  2gi. 

18.  When  5yd.  3qr.  lna.  of  cloth  cost  $  4,  how  much  cloth 
can  be  bought  for  $  1  ?     How  much  for  $  36  ? 

Ans.  52yd.  lqr.  lna. 

19.  If  11T.  3cwt.  2qr.  of  hay  be  sufficient  to  keep  4  horses 
7T3T  months,  how  much  will  keep  1  horse  the  same  time  ?  How 
much  23  horses?  Ans.  64T.  5cwt.  121b.  8oz. 

20.  If  12  men  can  dig  a  certain  ditch  in  286  days  4h.  33m., 
how  long  will  it  require  1  man  to  do  the  same  labor  ?  How 
long  72  men?  Ans.  47  days  16h.  45m.  30sec. 

21.  If  27yd.  lqr.  of  cloth  be  required  to  make  21  coats,  how 
many  yards  will  be  required  to  make  11  coats  ? 

22.  If  a  train  of  cars  move  at  the  rate  of  174m.  26rd.  in 
7  hours,  how  far  will  it  move  in  1  hour?  How  far  in  10 
hours?  Ans.  248m.  5fur.  20rd. 

23.  If  4  cases  of  shoes,  containing  60  pairs  each,  cost  $  192, 
what  will  1  pair  cost  ?     What  will  25  cases  cost  ? 

24.  When  3A.  2R.  20rd.  of  land  will  buy  4  hogsheads  of 
molasses,  how  much  land  will  buy  1  hogshead  ?  How  much 
30  hogsheads  ?  Ans.  27A.  OR.  30rd. 

25.  If  a  man  can  travel  20deg.  49m.  5fur.  35rd.  5yd.  3in.  in 
9  weeks,  how  far  would  he  travel  in  1  week  ?  How  far  in  90 
weeks  ?  Ans.  207deg.  13m.  lfur.  25rd.  5yd. 


PROPERTIES  OF  NUMBERS. 
DEFINITIONS. 

162.  An  integer  is  a  whole  number ;  as  1,  7,  16. 
All  whole  numbers  are  either  prime  or  composite. 

163.  A  prime  number  is  a  number  which  can  be  exactly 
divided  only  by  itself  and  1 ;  as  1,  3,  5,  7,  11. 

A  composite  number  is  a  number  which  can  be  exactly  di- 
vided by  some  number  besides  itself  and  1  ;  as  6,  9,  14,  18. 
12* 


138  PEOPERTIES   OF   NUMBERS. 

164.  A  factor  of  a  number  is  such  a  number  as  will,  by- 
being  taken  an  entire  number  of  times,  produce  it ;  as,  3  is  a 
factor  of  9,  and  4  a  factor  of  16. 

165.  A  prime  factor  of  a  number  is  a  prime  number  that 
will  exactly  divide  it ;  thus  the  prime  factors  of  10  are  the 
prime  numbers  1,  2,  and  5. 

Note.  —  Unity  or  1  is  not  generally  regarded  as  a  prime  factor,  since  mul- 
tiplying or  dividing  any  number  by  1  does  not  alter  its  value.  It  therefore 
will  be  omitted  when  speaking  of  the  prime  factors  of  numbers. 

A  composite  factor  of  a  number  is  a  composite  number 
that  will  exactly  divide  it ;  thus,  G  and  8  are  composite  factors 
of  48. 

166.  Numbers  are  prime  to  each  other  when  they  have 
no  factor  in  common ;    thus,  4,  9,  and  23  are  prime  to  each 

other. 

167.  An  aliquot  part  of  a  number  is  such  a  part  as  will 
exactly  divide  it;  as,  1,  3,  and  5  are  aliquot  parts  of  lo. 

Note.  —  The  aliquot  parts  of  a  number  include  all  its  factors,  prime  and 
composite. 

An  aliquant  part  of  a  number  is  such  a  part  as  will  not  ex- 
actly divide  it ;  as  2,  4,  5,  7,  and  8  are  aliquant  parts  of  9. 

168.  The  reciprocal  of  a  number  is  the  quotient  arising 
from  dividing  1  by  the  number ;  thus,  the  reciprocal  of  2  is  4-. 

169.  The  power  of  a  number  is  the  product  obtained  by 
taking  the  number  a  certain  number  of  times  as  a  factor ;  thus 
25  is  a  power  of  5. 

Note.  —  When  the  number  is  taken  once,  it  is  called  its  first  power;  when 
taken  twice,  as  a  factor,  the  product  is  called  its  second  power;  and  so  on. 
The  second  power  of  a  number  is  sometimes  termed  its  square,  and  the  third 
power,  its  cube. 

170.  The  exponent  of  a  power  is  a  figure  written  at  the 
right  of  a  number,  and  a  little  above  it,  to  show  how  many 
times  it  is  taken  as  a  factor ;  thus,  in  the  expression  42,  the 
exponent  is  the  2,  and  the  whole  is  read  4  second  power ;  and 
in  73,  it  is  the  3,  and  the  whole  read  7  third  power. 

Note.  —  The  first  power  of  a  number  being  always  the  number  itself,  its 
exponent  is  not  expressed. 


PROPERTIES   OP   NUMBERS.  139 


PROPERTIES  OF  PRIME  NUMBERS. 

171.  No  direct  process  of  detecting  prime  numbers  has  been 

discovered. 

Note.  —  A  few  facts,  such  as  are  given  below,  if  kept  in  mind,  will  aid 
somewhat  in  ascertaining  whether  a  number  is  prime  or  not. 

172.  The  only  even  prime  number  is  2  ;  since  all  other  even 
numbers,  as  4,  6,  8,  and  10,  it  is  evident,  can  be  exactly  divided 
by  2,  and  therefore  must  be  composite. 

173.  The  only  prime  number  having  5  for  a  unit  or  right- 
hand  figure  is  5  ;  since  every  other  whole  number  thus  termi- 
nating, as  15,  25,  35,  and  45,  can  be  exactly  divided  by  5,  and 
therefore  must  be  composite. 

174.  Every  prime  number,  except  2  and  5,  must  have  1,  3, 
7,  or  9  for  the  right-hand  figure  ;  since  all  other  numbers  are 
composite. 

175.  Every  prime  number  above  3,  when  divided  by  6,  must 
leave  1  or  5  for  a  remainder  ;  since  every  prime  number  above 
3  is  either  1  greater  or  1  less  than  6,  or  some  exact  number  of 
times  6. 

176.  In  a  series  of  odd  numbers  written  in  their  proper  or 
natural  order,  if  beginning  with  3  every  third  number,  with  5 
every  fifth,  with  7  every  seventh,  be  cancelled,  as  composite, 
the  remaining  numbers,  with  2,  will  be  the  prime  numbers  of 
the  natural  series.  Thus,  in  the  series  1,  3,  5,  7,  0,  11, 13,  X$, 
17,  19,  U,  23,  ft$,  n,  29,  31,  £3,  £&  37,  #0,  41,  43,  £$, 
47,  ^0,  every  third  number  from  the  3,  every  fifth  from  the  5, 
every  seventh  from  the  7,  every  ninth  from  the  9,  and  so  on, 
being  cancelled,  the  remaining  numbers,  with  2,  are  all  the 
prime  numbers  under  50. 

Note  1.  —  In  the  series,  every  third  number  from  the  3  contains  that  num- 
ber as  a  factor;  every  fifth  number  from  the  5,  that  number  as  a  factor;  and 
so  on. 

Note  2.  —  The  whole  number  of  prime  numbers  from  1  to  100,000  is  9,593. 
Although  all  of  these,  except  2  and  5,  end  in  1,  3,  7,  or  9,  there  are,  within 
the  same  range,  no  less  than  30,409  composite  numbers  terminating  with 
some  one  of  the  same  figures. 


140 


PROPERTIES   OF   NUMBERS. 


177.     All  the  prime  numbers  not  larger  than  4049  are  in- 
cluded in  the  following 


TABLE   OF  PRIME  NUMBERS. 


1 

233 

557 

883 

1249 

1607 

2003 

2389 

2791 

3229 

3631 

2 

239 

563 

887 

1259 

1609 

2011 

2393 

2797 

3251 

3637 

3 

241 

569 

907 

1277 

1613 

2017 

2399 

2801 

3253 

3643 

5 

251 

571 

911 

1279 

1619 

2027 

2411 

2803 

3257 

3659 

7 

257 

577 

919 

1283 

1621 

2029 

2417 

2819 

3259 

3671 

11 

263 

587 

929 

1289 

1627 

2039 

2423 

2833 

3271 

3673 

13 

269 

593 

937 

1291 

1637 

2053 

2437 

2837 

3299 

3677 

17 

271 

599 

941 

1297 

1657 

2063 

2441 

2843 

3301 

3691 

19 

277 

601 

947 

1301 

1663 

2069 

2447 

2851 

3307 

3697 

23 

281 

607 

953 

1303 

1667 

2081 

2459 

2857 

3313 

3701 

29 

283 

613 

967 

1307 

1669 

2083 

2467 

2861 

3319 

3709 

31 

293 

617 

971 

1319 

1693 

2087 

2473 

2879 

3323 

3719 

37 

307 

619 

977 

.  1321 

1697 

2089 

2477 

2887 

3329 

3727 

41 

311 

631 

983 

1327 

1699 

2099 

2503 

2897 

3331 

3733 

43 

313 

641 

991 

1361 

1709 

2111 

2521 

2903 

3343 

3739 

47 

317 

643 

997 

1367 

1721 

2113 

2531 

2909 

3347 

3761 

53 

331 

647 

1009 

1373 

1723 

2129 

2539 

2917 

3359 

3767 

59 

337 

653 

1013 

1381 

1733 

2131 

2543 

2927 

3361 

3769 

61 

347 

659 

1019 

1399 

1741 

2137 

2549 

2939 

3371 

3779 

67 

349 

661 

1021 

1409 

1747 

2141 

2551 

2953 

3373 

3793 

71 

353 

673 

1031 

1423 

1753 

2143 

2557 

2957 

3389 

3797 

73 

359 

677 

1033 

1427 

1759 

2153 

2579 

2963 

3391 

3803 

79 

367 

683 

1039 

1429 

1777 

2161 

2591 

2969 

3407 

3821 

83 

373 

691 

1049 

1433 

1783 

2179 

2593 

2971 

3413 

3823 

89 

379 

701 

1051 

1489 

1787 

2203 

2609 

2999 

3433 

3833 

97 

383 

709 

1061 

1447 

1789 

2207 

2617 

3001 

3449 

3847 

101 

389 

719 

1063 

1451 

1801 

2213 

2621 

3011 

3457 

3851 

103 

397 

727 

1069 

1453 

1811 

2221 

2633 

3019 

3461 

3853 

107 

401 

733 

1087 

1459 

1823 

2237 

2647 

3023 

3463 

3863 

109 

409 

739 

1091 

1471 

1831 

2239 

2657 

3037 

3467 

3877 

113 

419 

743 

1093 

1481 

1847 

2243 

2659 

3041 

3469 

3881 

127 

421 

751 

1097 

1483 

1861 

2251 

2663 

3049 

3491 

3889 

131 

431 

757 

1103 

1487 

1867 

2267 

2671 

3061 

3499 

3907 

137 

433 

761 

1109 

1489 

1871 

2269 

2677 

3067 

3511 

3911 

139 

439 

769 

1117 

1493 

1873  j 

227:1 

2683 

3079 

3517 

3917 

149 

443 

773 

1123 

1499 

1877' 

22  M 

2687 

3083 

3527 

3919 

151 

449 

787 

1129 

1511 

1879 

2287 

2689 

3089 

3529 

3923 

157 

457 

797 

1151 

1523 

1889 

2293 

2693 

3109 

3533 

3929 

163 

461 

809 

1153 

1527 

1901 

2297 

2699 

3119 

3539 

3931 

167 

463 

811 

1163 

1529 

1907 

2309 

2707 

3121 

3541 

3943 

173 

467 

821 

1171 

1533 

1913 

2311 

2211 

3137 

3547 

3947 

179 

479 

823 

1181 

1539 

1931 

2333 

2713 

3163 

3557 

3967 

181 

487 

827 

1187 

1547 

1933 

2339 

2719 

3167 

3559 

3989 

191 

491 

829 

1193 

1551 

1949 

2341 

2729 

3169 

3571 

4001 

193 

499 

839 

1201 

1553 

1951 

2347 

2731 

3181 

3581 

4003 

197 

503 

853 

1213 

1559 

1973 

2351 

2741 

3187 

3583 

4007 

199 

509 

857 

1217 

1571 

1979 

2357 

2749 

3191 

3593 

4013 

211 

521 

859 

1223 

1579 

1987 

2371 

2753 

3203 

3607 

4019 

223 

523 

863 

1229 

1583 

1993 

2377 

2767 

3209 

3613 

4021 

227 

541 

877 

1231 

1597 

1997 

2381 

2777 

3217 

3617 

4027 

229 

547 

881 

1237 

1601 

1999 

2383 

2789 

3221 

3623 

4049 

PROPERTIES   OP  NUMBERS.  141 


FACTORING. 

178.  Factoring  is  the  process  of  resolving  a  quantity 
into  its  factors. 

179.  Every  number  that  is  not  prime  is  composed  of  prime 
factors,  since  all  numbers  are  either  prime  or  composite  ;  and, 
if  composite,  can  be  separated  into  factors,  which,  if  themselves 
composite,  can  be  further  separated  into  those  that  shall  be 
prime. 

180.  To  resolve  a  composite  number  into  its  prime  factors. 
Ex.  1.  It  is  required  to  find  the  prime  factors  of  42. 

Ans.  2,  3,  7. 

operation.  *\y"e  divide  by  2,  the  least  prime  number  greater 

2  4  2  than  1,  and  obtain  the  quotient  21 ;  and,  since  21  is 

o  Try         a  composite  number,  we  divide  this  by  3,  and  obtain 
- —  for  a  quotient  7,  which  is  a  prime  number.      The 

7  several  divisors  and  the  last  quotient,  all  being  prime, 

constitute  all  the  prime  factors  of  42,  which,  multi- 
plied together,  they  equal.     Hence 

Divide  the  given  number  by  any  prime  number  that  will  exactly  divide 
it,  and  the  quotient,  if  a  composite  number,  in  the  same  manner;  and  so 
continue  dividing,  until  a  prime  number  is  obtained  for  a  quotient. 
The  several  divisors  and  the  last  quotient  will  be  the  prime  factors 
required. 

Note  1.  —  The  composite  factors  of  any  number  may  be  found  by  multiply- 
ing together  two  or  more  of  its  prime  factors. 

Note  2.  —  Such  prime  factors  as  two  or  more  numbers  may  have  alike,  are 
termed  prime  factors  common  to  them ;  and  these  may  be  readily  determined 
after  the  numbers  are  resolved  into  their  prime  factors. 

Examples. 

2.  What  are  the  prime  factors  of  105  ?  Ans.  3,  5,  7. 

3.  Resolve  220  into  its  prime  factors. 

4.  What  are  the  prime  factors  of  936  ? 

Ans.  2,  2,  2,  3,  3,  13. 

5.  What  are  the  prime  factors  of  1953  ? 

6.  Resolve  12462  into  its  prime  factors.     Ans.  2, 3, 31,  67. 

7.  Resolve  19987  into  its  prime  factors.       Ans.  11, 23, 79. 

8.  What  are  the  prime  factors  common  to  225,  435,  and 
540  ?  Ans.  3,  5. 


142  PROPERTIES    OF   NUMBERS. 

9.  What  are  the  prime  factors  common  to  960,  1568,  and 
5824? 

10.  What  are  the  prime  factors  common  to  2340,  11934, 
12987,  and  14859  ?  Ans.  3,  3,  13. 

11.  A  man  has  105  apples,  which  he  wishes  to  distribute  into 
small  parcels,  each  of  equal  numbers ;  what  are  the  smallest 
whole  numbers,  greater  than  1,  into  which  they  may  be  exactly 
divided  ?  Ans.  3,  5,  and  7. 

DIVISIBILITY  OF  NUMBERS. 

181.  One  number  is  said  to  be  divisible  by  another,  when 
the  latter  will  divide  the  former  without  a  remainder.  Thus, 
9  is  divisible  by  3. 

182.  One  number  is  divisible  by  another,  when  it  contains 
all  the  prime  factors  of  that  number.  Thus,  12,  which  contains 
all  the  factors  of  4,  is  divisible  by  4. 

183.  All  even  numbers,  or  such  as  terminate  with  0,  2,  4,  6, 
or  8,  are  divisible  by  2,  since  each  of  them  contains  2  as  a  fac- 
tor.    Thus,  10,  24,  36,  58,  are  each  divisible  by  2. 

184.  AU  numbers  which  terminate  with  0  or  5  are  divisible 
by  5,  since  each  of  them  contains  5  as  a  factor.  Thus,  20,  25, 
50,  are  each  divisible  by  5. 

185.  Every  number  is  divisible  by  4,  or  any  other  number 
that  will  exactly  divide  100,  when  its  two  right-hand  figures  are 
divisible  by  the  same*  For  any  figure  on  the  left  of  the  two 
right-hand  figures  must  express  one  or  more  hundreds,  and  a 
factor  of  one  hundred  is  a  factor  of  any  number  of  hundreds ; 
so,  if  the  sum  exactly  divides  the  units  and  tens  of  a  number, 
the  entire  number  will  be  divisible  by  it.  Thus,  116  is  divisi- 
ble by  4;  140,  by  20 ;  225,  by  25  ;  and  450,  by  50. 

186.  Every  number  is  divisible  by  8,  or  any  other  number 
that  will  exactly  divide  1000,  when  its  three  right-hand  figures 
are  divisible  by  the  same.  For  any  figure  on  the  left  of  the 
three  right-hand  figures  must  express  one  or  more  thousands, 
and  a  factor  of  one  thousand  is  a  factor  of  any  number  of 
thousands ;  so,  if  the  sum  exactly  divides  the  units,  tens,  and 


PROPERTIES   OP   NUMBERS.  143 

hundreds  of  a  number,  the  entire  number  will  be  divisible  by- 
it.  Thus,  1824  is  divisible  by  8  ;  1840,  by  40  ;  3375,  by  125  ; 
2750,  by  250 ;  and  4500,  by  500. 

187.  Every  number  the  sum  of  whose  digits  3  or  9  will 
exactly  divide,  is  divisible  by  3  or  9.  For  10,  or  any  power  of 
10,  less  1,  gives  a  number,  as  9,  99,  999,  &c.,  which  is  divisible 
by  3  and  by  9.  Hence,  any  number  of  tens,  hundreds, 
thousands,  &c,  less  as  many  units,  must  be  divisible  by  3  and 
by  9 ;  and  if  the  excess  of  units  denoted  by  the  significant 
figures,  in  the  aggregate,  is  likewise  divisible  by  3  and  by  9, 
it  follows  that  the  entire  number  is  thus  divisible.  For  exam- 
ple, 7542  is  a  number,  the  sum  of  whose  digits  is  divisible  by 
3  and  by  9 ;  and  separated  into  tens,  hundreds,  and  thousands, 
it  is  equal  to  7000  +  500  +  40  +  2.  Now,  7000  =  7  X 
1000  =  7  X  (999  +  1)  =  7  X  999  +  7 ;  500  =  5  X  100 
=  5  X  (99  + 1)  =  5  X  99  +  5  ;  and  40  =  4  X  10  =  4 
X  (9  +  1)  =  4  X  9  +  4.  Therefore,  7542  =  7  X  999  + 
5X  99 +  4X9  +  7  +  5  +  4  +  2.  The  remainders 
7  +  5  +  4  +  2,  corresponding  with  the  significant  figures 
of  the  number,  added  together,  equal  18,  which  sum  being 
divisible  by  3  and  by  9,  it  is  evident  that  7542  is  divisible  in 
like  manner. 

Note. — Upon  the  property  of  9  now  explained  depends  the  method  of 
proving,  by  excess  of  nines,  multiplication  (Art.  63),  and  division  (Art.  75). 

The  same  method  of  proof  may  be  resorted  to  in  addition 
and  in  subtraction.     Thus, 

To  prove  Addition.  Find  the  excess  of  nines  in  each  num- 
ber added,  and  then  the  excess  of  nines  in  the  sum  of  these 
results ;  which,  if  the  work  be  right,  will  equal  the  excess 
of  nines  in  the  answer. 

To  prove  Subtraction.  Find  the  excess  of  nines  in  the  sub- 
trahend, and  also  in  the  remainder,  and  then  the  excess  of 
nines  in  the  sum  of  these  results  ;  which,  if  the  work  be  right, 
will  equal  the  excess  of  nines  in  the  minuend. 

188i  Every  number  occupying  four  places,  in  which  two 
like  significant  figures  have  two  ciphers  between  them,  is  divisi- 
ble by  7,  11,  and  13.  Thus,  9009,  1001,  300300,  4004,  &c, 
are  each  divisible  by  7,  11,  and  13. 


144  PROPERTIES   OF   NUMBERS. 

189.  Every  number  is  divisible  by  11,  in  which  the  sum  of 
the  digits  in  the  odd  places  is  equal  to  the  sum  of  the  digits  in 
the  even  places,  or  in  which  the  difference  of  their  sums  can  be 
exactly  divided  by  11.  Thus,  8305,  in  which  3  +  5  =  8  +  0, 
and  628001,  in  which  2  +  0  +  1  and  6  +  8  +  0  differ  by 
11,  are  each  divisible  by  11. 

190.  Every  number  divisible  by  two\or  more  numbers,  which 
are  prime  to  each  other,  is  divisible  by  their  product  For, 
being  prime  to  each  other,  dividing"  by  one  of  the  numbers 
does  not  cancel  the  others  as  factors.  Thus,  770,  being  divisi- 
ble by  2,  5,  and  7,  which  are  prime  to  each  other,  is  divisible 
by  70,  their  product. 

191.  Every  even  number,  the  sum  of  whose  digits  6  wiU 
exactly  divide,  is  divisible  by  6.  For  being  even,  it  is  divisible 
by  2  (Art.  183),  and  its  digits  being  divisible  by  twice  3,  or  6, 
are  evidently  divisible  by  once  3,  so  that  the  number  is  also 
divisible  by  3 ;  and  as  the  2  and  the  3  are  prime  to  each  other, 
the  number  is  divisible  by  their  product,  or  6  (Art.  190). 
Thus,  174,  6312,  are  each  divisible  by  6. 

192.  Every  number  terminating  with  0  or  5  that  3  will 
exactly  divide,  is  divisible  by  15,  and  every  number  that  9  wiU 
exactly  divide,  is  divisible  by  45.  For,  terminating  with  0  or 
5,  it  is  divisible  by  5  (Art.  184),  and,  as  3  and  9  are  each 
prime  to  5,  if  it  can  be  exactly  divided  by  3  or  by  9,  it  must 
be  divisible  by  5  X  3  =  15,  or  by  5  X  9  =  45.  Thus,  75, 
which  3  will  exactly  divide,  is  divisible  by  15 ;  and  90,  which 
9  will  exactly  divide,  is  divisible  by  45. 

Divisors  or  Measures. 

193.  A  divisor  or  measure  of  a  number  is  any  number 
that  will  divide  it  without  a  remainder.  Thus,  3  is  a  divisor 
or  measure  of  6,  and  5  a  divisor  or  measure  of  10. 

194.  To  find  all  the  divisors  or  measures  of  a  number. 

Ex.  1.  Required  all  the  divisors  of  60. 

Ans.  1,  2,  3,  4,  5,  6,  10,  15,  20,  30,  and  60. 


OPERATION. 

60   rrr  2 

X  2  X  3  X  5. 

1        2 

4=2X2 

3        6 

12  =  2  X  2  X 

5     10 

20  =  2  X  2  X 

15     30 

60  =  2  X  2  X 

PROPERTIES   OF   NUMBERS.  145 

Resolving  the 
number  into  its 
prime  factors, 
we  find  60  = 
2X2X3X5. 
o  JNow,anynum- 

3X5.     ^er  *s  divisible 
by  1,  and  ev- 
Ans.  1,  2,  3,  4,  5,  6,  10,  12,  15,  20,  30,  and  60.     ery  composite 

number  by  its 
prime  factors,  and  every  product  these  factors  can  form.  Now  1, 
being  a  divisor  of  every  number,  is  a  divisor  of  60,  and,  since  2  en- 
ters twice  as  a  factor  into  60,  it  is  evident  that  2,  and  4  =  2  x  2, 
are  also  divisors  of  60.  These  divisors  we  arrange  on  a  horizontal 
line,  and  determine  the  other  divisors  by  multiplying  those  on  this 
line  by  the  factor  3,  for  the  second  line  of  divisors,  by  5  for  the  third 
line,  and  by  3  x  5  for  the  fourth  line,  and  thus  obtain  all  the  pos- 
sible divisors  of  the  given  number. 

The  whole  number  of  divisors  is  12,  which  corresponds  to  the  pro- 
duct arising  from  multiplying  together  the  exponents,  each  increased 
by  1,  of  the  different  prime  factors  of  60  ;  thus,  of  the  different  prime 
factors,  since  2  enters  twice,  its  exponent  is  2,  -\-  1  =  3  ;  3  enters 
once,  its  exponent  is  1,  -j-  1  =  2  ;  5  enters  once,  its  exponent  is 
15  _|_  l  =  2 ;  and  3  x  2  X  2  =  12,  the  number  of  divisors.  The 
same  holds  true  in  all  cases.     Hence,  in  any  composite  number, 

To  find  the  Number  of  Divisors.  —  Multiply  together  the  ex- 
ponents, each  increased  by  1 ,  of  the  different  prime  factors  of  the  given 
number,  and  the  product  ivill  be  the  number  of  divisors  required.     And 

To  find  the  several  Divisors. — Form  from  the  prime  factors 
of  the  number  all  the  products  possible,  and  these  factoids  (including  1) 
and  products  will  be  the  divisors  required. 

Examples. 

2.  What  are  the  divisors  of  72  ? 

Ans.  1,  2,  3,  4,  6,  8,  9,  12,  18,  24,  36,  and  72. 

3.  Required  the  divisors  of  105  ? 

4.  How  many  divisors  has  1764  ?  Ans.  27. 

5.  How  many  divisors  has  3528  ? 

6.  How  many  divisors  has  5880  ?  Ans.  48. 

Common  Divisors  or  Measures. 

195.     A  common  divisor  or  measure  of  two  or  more  num- 
bers is  any  number  that  will  divide  them  without  a  remainder ; 
thus,  2  is  a  common  divisor  of  4,  8,  and  10. 
N    13 


146  PROPERTIES   OP   NUMBERS. 

196.  A  common  divisor  of  two  numbers  is  a  divisor  of  their 
sum,  and  also  of  their  difference.  Thus,  6,  a  common  divisor 
of  12  and  18,  is  a  divisor  of  their  sum,  40,  and  of  their  dif- 
ference, 6. 

197 #  A  common  divisor  of  the  remainder  and  the  divisor  is 
a  divisor  of  the  dividend.  Thus,  in  a  division  having  8  for  a 
remainder,  16  for  divisor,  and  24  for  dividend,  8,  a  common  di- 
visor of  8  and  1G,  is  also  a  divisor  of  the  24. 

198.  To  find  all  the  divisors  common  to  two  or  more  num- 
bers. 

Ex.  1.     Required  all  the  common  divisors  of  45  and  135. 

Ans.  1,  3,  5,  9,  15,  and  45. 
operation.  Resolving  the  given 

numbers      into     their 
5#  prime  factors^  we  find 

they  have  of  these  3, 
3,  and  5  in  common, 
and  these  common" 
X  5.  I)rime  factors  tcith  1, 
and  all  the  product*  ire 
are  able  to  form  from 
them  (Art.  194),  give 
all  the  common  divisors  required.  When  only  the  number  of  com- 
mon divisors  is  required,  it  may  readily  be  found  by  multiplying  to- 
gether the  exponents,  each  increased  by  I,  of  the  different  common 
prime  factors.     (Art.  194.) 

Examples. 

2.  What  are  the  common  divisors  of  51,  153,  and  255  ? 

3.  Required  the  several  common  divisors  of  180  and  360. 
Ans.  1,  2,  3,  4,  5,  6,  9,  10,  12,  15,  18,  20,  30,  36,  45,  GO, 

90,  and  180. 

4.  How  many  common  divisors  have  2025,  6075,  and  8100? 

Ans.  15. 

5.  How  many  common  divisors  have  4500  and  9000  ? 

Ans.  36. 

The  Greatest  Common  Divisor  or  Measure. 

199.  The  greatest  common  divisor  or  measure  of  two  or 
more  numbers  is  the  greatest  number  that  will  divide  each  of 


4  5 
1  3  5 

=  3 

=  3 

X  3 
X  3 

X  5. 
X  3   X 

Common  f  1 

Divisors   (  5 

Ans. 

3 
15 

1,3, 

9=3X3 
45=3X3 

5,  9,  15,  and  45, 

PROPERTIES   OF   NUMBERS.  147 

them  without  a  remainder.     Thus,  4  is  the  greatest  common 
divisor  of  8,  12,  and  16. 

200.  To  find  the  greatest  common  divisor  of  two  or  more 
numbers. 

Ex.  1.  Required  the  greatest  common  divisor  or  measure 
of  24  and  88.  Ans.  8. 

first  operation.  Resolving  the  numbers  into  their 

24  =  2X2  X^X      3       prime  factors,  thus,  24  =  2  X  2X2 

XQ—9V9V9V   11        X3,and88  =  2x2X2XH  = 

9         9         9  _  o        a  88'  We  find  the  fact0rS  2  X   2  x   2 

^  X  2  X  2  —  8.     Ans.     are   common  to  both.     Since  only 

these  common  factors,  or  the  pro- 
duct of  two  or  more  of  such  factors,  will  exactly  divide  both  numbers, 
it  follows  that  the  product  of  all  their  common  prime  factors  must  be  the 
greatest  factor  that  will  exactly  divide  both  of  them.  Therefore,  2x2 
X  2  =  8,  the  greatest  common  divisor  required. 

The  same  result  may  be  obtained  by  a  sort  of  trial  process,  as  by 
the  second  operation. 

It  is  evident,  since  24  cannot  be  ex- 

second  operation.  actly  divided  by  a  number  greater  than 

2  4)88(3  itself,  if  it  will  also  exactly  divide  88, 

7  2  it  will  be  the  greatest  common  divisor 

sought.     But,  on  trial,  we  find  24  will 

16)24(1  not  exactly  divide  88,  there  being  a  re- 

1  6  mainder,  16.     Therefore    24   is  not  a 
— o~\  i  n  /  9         common  divisor  of  the  two  numbers. 

/         1  We  know  that  a  common  divisor  of 

16  16  and  24  will,  also,  be  a  common  di- 

visor of  88  (Art.  197).  We  next  try  to 
find  that  divisor.  It  cannot  be  greater  than  16.  But  16  will  not  ex- 
actly divide  24,  there  being  a  remainder,  8;  therefore  16  is  not  the 
greatest  common  divisor. 

As  before,  the  common  divisor  of  8  and  16  will  be  the  common  di- 
visor of  24  and  88  (Art.  197)  ;  we  make  trial  to  find  that  divisor, 
knowing  that  it  cannot  be  greater  than  8,  and  find  8  will  exactly 
divide  16.     Therefore  8  is  the  greatest  common  divisor  required. 

The  last  method  may  be  often  contracted,  if 

third  operation.      there  should  be  observed  to  be  any  prime  factor 

24)88(3  in  a  remainder  which  is  not  common  to  the  pre- 

7  2  ceding  divisor,  by  cancelling  said  factor.    Thus, 

in  the  third  operation,  the  remainder  16  being 

<£  0  observed  to  contain  the  prime  factor  2,  common 

8)24(3     to  the  preceding  divisor  24,  we  cancel  it,  and, 

2  a  having  left  the  composite  factor  8,  we  divide  24 
by  that  factor.    There  being  no  remainder,  8  is 

the  greatest  common  divisor,  as  before  obtained. 


148  PROPERTIES   OF   NUMBERS. 

Rule  1.  —  Resolve  the  given  numbers  into  their  prime  factors.  The 
product  of  all  the  factors  common  to  the  several  numbers  will  be  the 
greatest  common  divisor.     Or, 

Rule  2.  —  Divide  the  greater  number  by  the  less,  and  if  there  be  a 
remainder  divide  the  preceding  divisor  by  it,  and  so  continue  dividing 
until  nothing  remains.  The  last  divisor  will  be  the  greatest  common  di- 
visor. 

Note  1.  —  When  the  greatest  common  divisor  is  required  of  more  than  two 
numbers,  find  it  of  two  of  them,  and  then  of  that  common  divisor  and  of  one  of 
the  other  numbers,  and  so  on  for  all  the  given  numbers.  The  last  common  di- 
visor will  be  the  greatest  common  divisor  required. 

Note  2.  —  If  any  remainder  be  such  a  prime  number  as  will  not  exactly 
divide  the  preceding  divisor,  or  if  the  last  remainder  be  1,  the  numbers,  being 
prime  to  each  other,  will  have  no  common  divisor. 

Examples. 

2.  What  is  the  greatest  common  divisor  of  56  and  1G8  ? 

3.  What  is  the  greatest  common  divisor  of  OG  and  128  ? 

4.  What    is    the    greatest   common    measure   of  57    and 
285  ?  Ans.  57. 

5.  What  is  the  greatest  common  measure  of  1G9  and  175  ? 


G.  What 

is 

the 

greatest   common   measure 

of    175    and 

455? 

Ans.  35. 

7.  What 

is 

the 

greatest    common    divisor 

of    1 69    and 

8GG? 

Ans.  1. 

8.  What 

is 

the 

greatest   common   measure 

of    47    and 

478? 

Ans.  1. 

9.  What 

is 

the 

greatest   common   measure 

of    84    and 

10G8? 

Ans.  12. 

10.  What 

is 

the 

greatest    common    divisor 

of    75    and 

1G5? 

Ans.  15. 

11.  What  is  the  greatest  common  measure  of  78,  234,  and 
4G8  ?  Ans.  78. 

12.  I  have  three  fields ;  one  containing  16  acres ;  the  second, 
20  acres ;  and  the  third,  24  acres.  Required  the  largest-sized 
lots,  containing  each  an  exact  number  of  acres,  into  which  the 
whole  can  be  divided.  Ans.  4  acre  lots. 

13.  A  farmer  has  12  bushels  of  oats,  18  bushels  of  rye,  24 
bushels  of  corn,  and  30  bushels  of  wheat.  Required  the 
largest  bins,  of  uniform  size,  and  containing  an  exact  number 
of  bushels,  into  which  the  whole  can  be  put,  each  kind  by 
itself,  and  all  the  bins  be  full. 


PROPERTIES   OF   NUMBERS.  149 


LEAST   COMMON  MULTIPLE. 

201  •  A  common  multiple  of  two  or  more  numbers  is  a 
number  that  can  be  divided  by  each  of  them  without  a  remain- 
der ;  thus,  14  is  a  common  multiple  of  2  and  7. 

The  least  common  multiple  of  two  or  more  numbers  is  the 
least  number  that  can  be  divided  by  each  of  them  without  a 
remainder ;  thus,  12  is  the  least  common  multiple  of  4  and  6. 

2®2t  A  multiple  of  a  number  contains  all  the  prime  factors 
of  that  number;  the  common  multiple  of  two  or  more  num- 
bers contains  all  the  prime  factors  of  each  of  the  numbers ; 
and  the  least  common  multiple  of  two  or  more  numbers  con- 
tains only  each  prime  factor  taken  the  greatest  number  of 
times  it  is  found  in  any  of  the  several  numbers.     Hence, 

1.  The  least  common  multiple  of  two  or  more  numbers  must 
be  the  least  number  that  will  contain  all  the  prime  factors  of 
them,  and  none  others. 

2.  The  least  common  multiple  of  two  or  more  numbers, 
which  are  prime  to  each  other,  must  equal  their  product. 

3.  The  least  common  multiple  of  two  or  more  numbers  must 
equal  the  product  of  their  greatest  common  divisor,  by  the 
factors  of  each  number  not  common  to  all  the  numbers. 

4.  The  least  common  multiple  of  two  or  more  numbers,  di- 
vided by  any  one  of  them,  must  equal  the  product  of  those 
factors  of  the  others  not  common  to  the  divisor. 

203.  To  find  the  least  common  multiple  of  two  or  more 
numbers. 

Ex.  1.  What  is  the  least  common  multiple  of  8,   16,  24, 

32,  44.  Ans.  1056. 

first  operation.  Resolving    the 

8  =  2  X  2  X      2  numbers  into  their 

16  =  2X2X      2X2  prime  factors,  we 

24  =  2X2  X     2X3  ^nci  the^r   differ- 

3  2  =  2  X  2  X     2X2X2  ent  Prime  factors 

4  4  =  2X2X11  *  bL2'   3'  ?nc* 
^  11.    The  greatest 

2X2X2X2X2X3X11=  1056  Ans.  number  of  times 

the  2  occurs  as  a 

factor  in  any  of  the  given  numbers  is  5  times ;  the  greatest  num- 

13* 


2)8 

SECOND    OPERATION. 

16     24     32 

44 

2)4 

8 

12     16 

22 

2)2 

4 

6        8 

11 

2)1 

2 

3        4 

11 

150  PROPERTIES   OF   NUMBERS. 

ber  of  times  3  occurs  in  any  of  the  numbers  is  once;  and  the 
greatest  number  of  times  the  1 1  occurs  in  any  of  the  numbers  is 
once.  Hence,  2,  2,  2,  2,  2,  3,  and  11  must  be  all  the  prime  factors 
necessary  in  composing  8,  16,  24,  32,  and  44;  and  consequently, 
1056,  the  product  of  these  factors,  is  the  least  common  multiple  re- 
quired (Art.  202). 

Having  arranged  the 
numbers  on  a  horizon- 
tal line,  we  divide  by 
2,  a  prime  number  that 
will  divide  two  or  more 
of  them  without  a  re- 
mainder, and  write  the 
o         «  quotients  in  a  line  be- 

~  ~~n  low  ;  and  we  continue 

2X2X2X2X2X3XH  ==10o6  Ans.      to  divide  by  a  prime 

number  as  before,  till 
the  divisor  and  remainders  are  all  prime  to  each  other.  Then,  these, 
since  they  include  all  the  factors  necessary  to  form  the  given  num- 
bers and  no  others,  we  multiply  together  for  the  required  least  com- 
mon multiple,  and  obtain  1056,  as  before. 

The  least  common  multiple  of  two  or  more  numbers  may  be  found 
generally  by  a  process  much  shorter  than  either  of  the  above  meth- 
ods, by  cancelling  any  number  that  is  a  factor  of  any  other  of  the 
given  numbers,  and  also  by  dividing  the  numbers  by  such  a  com- 
posite number  as  may  be  observed  to  be  their  common  or  greatest 
common  divisor. 

third  operation.  Thus,  in  the  third  operation, 

4  )  $     X  0     24     32     44  8  being  a  factor  of  several  of 

— o~\~ a o — tTT  ^e  num^)ers»  an(l  16  being  a 

w  '  oil  factor  of  one  other  number,  we 

3         4     11  cancel    them ;    and    observing 

that  4  is  the  greatest  common 
4X2X^X^X11  =  10ob  Ans.  divisor  of  the  remaining  num- 
bers, we  divide  them  by  it. 
We  next  divide  by  2,  as  in  the  second  operation.  The  numbers  in 
the  lower  line  then  being  prime  to  each  other,  we  multiply  them  and 
the  divisors  together,  and  obtain  1056  as  the  least  common  multiple. 

fourth  operation.  The  fourth  operation  exhibits 

$     X  0     24132     44  a  process   yet   more  contracted. 

— ~ — t-i  The  8  and  16  being  factors  each 

of  one  or  more  of  the  other  num- 

24X8X11  =  1056  Ans.     bers,  we  cancel  them,  as  in  the 

third  operation.  Of  the  remain- 
ing numbers  we  cut  off  24  by  a  short  vertical  line  from  the  rest  as  a 
factor  of  the  least  common  multiple  sought.  We  then  strike  out  of 
the  two  remaining  numbers  the  largest  factor  each  has  in  common 
with  the  24,  by  dividing  each  of  them  by  the  greatest  common  divisor 


PROPERTIES   OF   NUMBERS.  151 

between  it  and  24,  and  write  the  result  beneath.  The  numbers  in 
the  lower  line  having  no  factor  in  common,  we  carry  the  process  no 
further.  The  continued  product  of  the  number  cut  off  by  the  num- 
bers in  the  lower  line  gives  1056,  the  least  common  multiple,  as  by 
the  other  methods.  In  this  instance  we  cut  off  the  24,  but  either 
the  32  could  have  been  separated  from  the  rest,  or  the  44  cut  off, 
and  the  needless  factors  striken  out  with  like  result.  If,  however, 
we  had  cut  off  the  44,  the  numbers  placed  in  the  second  line  would 
have  contained  factors  common  to  each  other,  so  that  it  would  have 
been  necessary  in  that  line  to  have  cut  off  and  stricken  out  factors 
as  before.  The  reason  for  this  abridged  process  is,  that  by  the  sepa- 
rating off,  and  by  the  striking  out  of  factors,  we  get  rid,  in  an  expe- 
ditious way,  of  the  factors  not  required  to  form  the  least  common 
multiple  sought. 

Rule  1.  —  Resolve  the  given  numbers  into  their  prime  factors.  The 
product  of  these  factors,  taking  each  factor  only  the  greatest  number  of 
times  it  occurs  in  any  of  the  numbers,  will  be  the  least  common  multiple. 
Or, 

Rule  2.  —  Having  arranged  the  numbers  on  a  horizontal  line,  can- 
cel such  of  them  as  are  factors  of  any  of  the  others,  and  separate  some 
convenient  one  from  the  rest.  Reject  from  each  of  the  numbers  remain- 
ing the  greatest  factor  common  to  it  and  that  number,  and  write  the 
result  in  a  line  below.  Should  there  be  in  the  second  line  numbers  hav- 
ing factoids  in  common,  proceed  as  before  ;  and  so  continue  until  the 
numbers  written  below  are  prime  to  each  other.  The  continued  product 
of  the  number  or  numbers  separated  from  the  others  with  those  in  the  last 
line  will  be  the  least  common  multiple. 

Note  1.  —  Some  give  a  preference  to  the  following  rule  for  finding  the  least 
common  multiple:  Having  arranged  the  numbers  on  a  horizontal  line,  divide  by 
such  a  prime  number  as  will  exactly  divide  two  or  more  of  them,  and  write  the 
quotients  and  undivided  numbers  in  a  line  beneath.  So  continue  to  divide  until  the 
quotients  shall  be  prime  to  each  other.  Then  the  product  of  the  divisors  and  the 
numbers  of  the  last  line  will  be  the  least  common  multiple. 

Note  2.  —  The  least  common  multiple  of  two  or  more  numbers  that  are 
prime  to  each  other  is  found  by  multiplying  them  together  (Art.  202). 

Note  3.  —  When  a  single  number  alone  is  prime  to  all  the  rest,  it  may  be 
separated  off,  and  used  only  as  a  factor  of  the  least  common  multiple  sought. 

Note  4.  —  When  the  least  common  multiple  of  several  numbers,  and  all  the 
numbers  except  one,  which  is  prime  to  the  others,  are  given,  to  find  the  un- 
known number,  divide  the  least  common  multiple  given  by  that  of  the  known 
numbers  (Art.  202). 

Examples. 

2.  What  is  the  least  common  multiple  of  3,  13,  37,  and 
91. 

3.  What  is  the  least  common  multiple  of  9,  14,  30,  35, 
and  47  ?  Ans.  29610. 


152  PROPERTIES   OF   NUMBERS. 

OPERATION. 

9     14     3  0  |  3  5     4  7. 

9       ft\     6 


3 
47  X  35  X  0  X  3  =  29610  Ans. 

4.  What  is  the  least  common  multiple  of  6,  8,  10,  18,  20, 
and  24  ? 

5.  What  is  the  least  common  multiple  of  14,  19,  38,  and 
57  ?  Ans.  798. 

6.  What  is  the  least  common  multiple  of  20,  36,  48,  and 
50  ?  Ans.  3600. 

7.  What  is  the  least  common  multiple  of  15,  25,  35,  45,  and 
100  ?  Ans.  6300. 

8.  What  is  the  least  common  multiple  of  100,  200,  300,  400, 
and  575  ?  Ans.  27600. 

9.  The  least  common  multiple  of  1,  2,  3,  4,  5,  6,  8,  9,  and 
one  other  number  prime  to  them,  is  2520.  What  is  that  other 
number  ?  Ans.  7. 

10.  What  is  the  least  common  multiple  of  18,  24,  36,  126, 
20,  and  48  ? 

11.  I  have  four  different  measures;  the  first  contains  4 
quarts,  the  second  6  quarts,  the  third  10  quarts,  and  the  fourth  12 
quarts.  How  large  is  a  vessel,  that  may  be  filled  by  each  one  of 
these,  taken  a  certain  number  of  times  full  ?     Ans.  60  quarts. 

12.  What  is  the  smallest  sum  of  money  with  which  I  can 
purchase  a  number  of  oxen  at  $  50  each,  cows  at  $  40  each,  or 
horses  at  $  75  each  ?  Ans.  §  600. 

MISCELLANEOUS  EXAMPLES. 

1.  How  many  times  does  7  occur  as  a  factor  of  6174  ? 

Ans.  3  times. 

2.  Required  the  largest  prime  factor  of  5775. 

3.  Required  the  largest  composite  factor  of  19929/ 

Ans.  6643. 

4.  Required  the  quotients  of  2338  divided  by  its  two  prime 
factors  next  larger  than  1.  Ans.  1169  ;  334. 

5.  Required  all  the  prime  numbers  that  will  divide  17385 
without  a  remainder. 


PROPERTIES  OF  NUMBERS.  153 

6.  A  farmer  has  3000  bushels  of  grain  ;  which  are  the  three 
smallest-sized  bags,  and  the  three  largest-sized  bins,  holding  an 
exact  number  of  bushels,  that  will  each  measure  the  same 
without  a  remainder  ? 

Ans.  Bags  of  1,  2,  or  3  bushels  each;  and  bins  of  1500, 
1000,  or  750  bushels  each. 

7.  A  teacher  having  a  school  consisting  of  152  ladies  and 
136  gentlemen,  divided  it  in  such  a  manner  that  each  class  of 
ladies  equalled  each  class  of  gentlemen,  and  the  classes  were 
the  largest  the  school  would  admit  of,  and  have  them  all  of  the 
same  size.  Required  the  number  of  classes,  and  the  number 
in  each  class. 

Ans.  19  classes  of  ladies,  17  classes  of  gentlemen,  and  8 
pupils  in  a  class. 

8.  At  noon  the  second,  minute,  and  hour  hand  of  a  clock 
are  together ;  how  long  after  will  they  be  again,  for  the  first 
time,  in  the  same  position  ? 

9.  J.  Porter  has  a  four-sided  garden,  the  first  side  of  which 
is  348  feet  in  length ;  the  second,  372  feet ;  the  third,  444  feet ; 
and  the  fourth,  492  feet.  Required  the  length  of  the  longest 
rails  that  can  be  used  in  fencing  it,  allowing  the  end  of  each 
rail  to  lap  by  the  other  9  inches,  and  all  the  panels  to  be  of 
equal  length ;  also,  the  number  of  rails,  if  5  rails  be  allowed  to 
each  panel.  Ans.     Length  12ft.  9in. ;  and  690  rails. 

10.  L.  Ford  has  5  pieces  of  land,  the  first  containing  3 A. 
2R.  lp.;  the  second,  5A.  3R.  15p.;  the  third,  8A.  29p. ;  the 
fourth,  12A.  3R.  17p.;  and  the  fifth,  15A.  31p.  Required  the 
largest  sized  house-lots,  containing  each  an  exact  number  of 
square  rods,  into  which  the  whole  can  be  divided. 

Ans.  1A.  27p.  each. 

11.  What  three  numbers  between  30  and  140  have  12  for 
their  greatest  common  divisor,  and  2772  for  their  least  common 
multiple.  Ans.  36,  84,  and  132. 

12.  Four  men,  A,  B,  C,  and  D,  are  engaged  in  making  reg- 
ular excursions  into  the  country,  between  which  each  stays  at 
home  just  1  day ;  and  A  is  always  absent  exactly  3  days,  B  5 
days,  and  C  and  D  7  days.  Provided  they  all  start  off  on  the 
same  day,  how  many  days  must  elapse  before  they  can  all  be 
at  home  again  on  the  same  day  ?  Ans.  23  days. 


154  COMMON   FRACTIONS. 


COMMON   FRACTIONS. 


204.  A  fraction  is  an  expression  denoting  one  or  more 
equal  parts  of  a  unit. 

205.  A  fractional  unit  is  one  of  the  equal  parts  into  which 
the  whole  thing  or  integral  unit  has  been  divided.  Thus  halves, 
thirds,  &c,  being  equal  parts  of  integral  units  or  whole  things, 
are  fractional  units. 

206.  The  unit  of  a  fraction  is  the  unit  or  whole  thing  from 
which  its  fractional  parts  have  been  derived. 

207.  A  Common  Fraction  is  expressed  by  two  numbers 
one  above  the  other,  with  a  line  between  them. 

208*  The  number  below  the  line  is  called  the  denominator. 
It  shows  into  how  many  parts  the  whole  number  has  been 
divided.  It  gives  name  to  the  fraction  and  value  to  the  frac- 
tional unit.  Thus,  in  the  expression  f ,  the  denominator  is  7, 
indicating  that  the  unit  of  the  fraction  has  been  divided  into  7 
equal  parts,  and  that  the  value  of  the  fractional  unit  is  one 
seventh. 

The  number  above  the  line  is  called  the  numerator.  It 
shows  how  many  parts  have  been  taken,  or  numbers  the  frac- 
tional units  expressed  by  the  fraction.  Thus,  in  the  expression 
f ,  the  numerator  is  2,  indicating  that  the  fractional  unit,  which 
is  one  seventh,  has  been  taken  2  times. 

209*  The  terms  of  a  fraction  are  its  numerator  and  denomi- 
nator. Thus,  the  terms  of  the  fraction  f  are  the  numerator  2 
and  the  denominator  3. 

210.  A  proper  fraction  is  one  whose  numerator  is  less  than 
the  denominator ;  as  f ,  §,  J. 

211.  An  improper  fraction  is  one  whose  numerator  is  equal 
to,  or  greater  than,  the  denominator ;  as  JJ-,  fy9  -\5-. 

212.  A  mixed  number  is  a  whole  number  with  a  fraction  ; 
as  3£,  16 J,  90 £. 

213.  A  simple  or  single  fraction  has  but  one  numerator  and 
one  denominator.     It  may  be  either  proper  or  improper;  as 


COMMON    FRACTIONS.  155 

214*  A  compound  fraction  is  a  fraction  of  a  fraction,  or  two 
or  more  fractions  connected  by  the  word  of;  as  J-  of  J  of  ^, 
J  off  of  $. 

215  •  A  complex  fraction  is  a  fraction  having  a  fraction  or  a 
mixed  number  for  its  numerator  or  denominator,  or  both ;  as 

f  h'  13'  h' 

216.  A  fraction  is  an  expression  of  division  ;  the  numer- 
ator answering  to  the  dividend,  and  the  denominator  to  the 
divisor  (Art.  67) ;  and  the  value  of  a  fraction  is  the  quotient 
arising  from  the  division  of  the  numerator  by  the  denominator 
(Art.  80).  Thus,  in  the  fraction  JT5-,  the  numerator  15  is  the 
dividend,  the  denominator  7  is  the  divisor,  and  the  value  ex- 
pressed 2 .J,  or  the  quotient  arising  from  the  division  of  the  15 
by  the  7. 

217.  Since  a  fraction  is  an  expression  of  division,  it  follows, 

1.  That,  if  the  numerator  be  multiplied,  or  the  denominator 
be  divided,  by  any  number,  the  fraction  is  multiplied  by  the  same 
number  (Art.  81). 

2.  TJiat,  if  the  numerator  be  divided,  or  the  denominator  mul- 
tiplied, by  any  number,  the  fraction  is  divided  by  the  same  num- 
ber (Art.  82). 

3.  That,  if  the  numerator  and  denominator  be  both  multiplied, 
or  both  divided,  by  the  same  number,  the  fraction  will  not  be 
changed  in  value  (Art.  83). 

REDUCTION  OF  COMMON  FRACTIONS. 

218.  Reduction  of  fractions  is  the  process  of  changing  their 
form  of  expression  without  altering  their  value. 

219.  A  fraction  is  in  its  lowest  terms,  when  its  numerator 
and  denominator  are  prime  to  each  other  (Art.  166). 

220.  To  reduce  a  fraction  to  its  lowest  terms. 

Ex.  1.  Reduce  J-|  to  its  lowest  terms.  Ans.  -£. 

birst  operation.  By  dividing  both  terms  of  the  fraction  by 

4  )  ft  ==  T2"  4,  a  factor  common  to  them  both,  it  is  re- 

4  )  tV  =  i     ^ns.     duced  to  &.     Dividing  both  terms  of  j%  by 

4,  a  factor  common  to  them  both,  it  is  re- 


156  COMMON   FRACTIONS. 

duced  to  £.     Now,  as  1  and  3  are  prime  to  each  other,  the  fraction  £ 
is  in  its  lowest  terms. 

The  same  result  is  often  more  readily  ob- 
second  operation.  tained  by  dividing  the  terms  of  the  fraction 

1^  )  if  ==  "3"  Ans.     by  their  greatest  common  divisor,  as  by  the 

second  operation. 
Since,  by  dividing  the  numerator  and  denominator  by  the  same 
number,  we  cancel  equal  factors  in  both,  and  diminish  them  in  the 
same  ratio,  their  relation  to  each  other  evidently  is  not  changed,  and 
the  value  of  the  fraction  remains  the  same  (Art.  217). 

Rule. — Divide  the  numerator  and  denominator  by  any  number 
greater  than  1  that  tvill  divide  them  both  ivithout  a  remainder,  and  thus 
proceed  until  they  are  prime  to  each  other.     Or, 

Divide  both  the  numerator  and  denominator  by  their  greatest  common 
divisor. 

Examples. 

2.  Reduce  £f  to  its  lowest  terms.  Ans.  54T. 

3.  Reduce  ||  to  its  lowest  terms.  Ans.  £ . 

4.  Reduce  $  £  to  its  lowest  terms.  Ans.  f . 

5.  Reduce  y1^  to  its  lowest  terms. 

6.  Reduce  -jV^  to  its  lowest  terms.  Ans.  ^. 

7.  Reduce  ££f  to  its  lowest  terms.  Ans.  ^j. 

8.  Reduce  a^Vg  to  its  lowest  terms. 

9.  Reduce  fff  to  its  lowest  terms.  Ans.  |||. 

10.  Reduce  tVtV  to  *ts  lowest  terms.  Ans.  tYtV 

11.  Reduce  §-f|  to  its  lowest  terms. 

12.  Reduce  TVrV  t0  *ts  lowest  terms.  Ans.  f  g|. 

13.  Reduce  £§  g  to  its  lowest  terms.  Ans.  ffe. 

221  #  To  reduce  an  improper  fraction  to  an  equivalent 
whole  or  mixed  number. 

Ex.  1.  How  many  yards  in  -^V/-  of  a  yard  ?         Ans.  6T3F. 

Since  19  nineteenths  make  one  yard, 

operation.  it   is   evident  there  will  be  as  many 

1  9  )  1  1  7  (  6T3g-  Ans.      yards  in  117  nineteenths  as  19  is  con- 

114  tained  times  in  117,  which  is  6 A  times. 

o  Therefore,  6^  yards  is  the  answer  re- 

**  quired. 

Rule.  —  Divide  the  numerator  by  the  denominator. 

Note.  —  Should  there  a  remainder  occur,  write  it  over  the  denominator,  and 
make  this  fraction  a  part  of  the  answer. 


COMMON   FRACTIONS.  157 

Examples. 

2.  Reduce  iff-  to  a  mixed  number.  Ans.  11T25. 

3.  Reduce  VtV"  to  a  mixea*  number.  Ans.  14Tf  F. 

4.  Reduce  -^j1-  to  a  mixed  number. 

5.  Reduce  f  ^{  to  a  mixed  number.  Ans.  3J-ff . 

6.  Change  liV°--  to  a  mixed  number.  Ans.  1 1 J  J. 

7.  Change  -£§-  to  a  mixed  number  Ans.  9 Iff. 

8.  Change  -*f  *  to  a  whole  number.  Ans.  125. 

9.  Change  -|f  to  a  whole  number. 

222.  To  reduce  a  whole  or  mixed  number  to  an  improper 
fraction. 

Ex.  1.  Reduce  19  to  a  fraction  whose  denominator  shall 
be  7. 

operation.  Since  there  are    7  sevenths  in  1 

19   X  7  =  133.  whole   one,    19    whole   ones  =  133 

133  sevenths  ±=  *f* ,  Ans.    s^enths  =  if*. 

2.  Reduce  17f  to  an  improper  fraction  ?  Ans.  -858-. 

OPERATION. 

5  Since  there  are  5  fifths  in  1  whole  one, 

«*  in  17  whole  ones  there  are  85  fifths,  and 

o  £  adding  3  fifths  for  the  fraction,  we  have  8| 

as  the  equivalent  of  1 7|.     Hence  the 
o 

8  8  fifths  ==  *£,  Ans. 

Rule.  —  Multiply  the  whole  number  by  the  given  denominator,  and 
to  the  product  add  the  numerator  of  the  fractional  part,  if  any ;  and 
write  the  result  over  the  denominator. 

Note.  —  A  whole  number  may  be  expressed  in  its  simplest  fractional  form, 
by  taking  it  for  a  numerator  with  Ifor  a  denominator.  Thus,  4  may  be  written 
4 ,  and  read  4  ones. 

Examples. 

3.  Reduce  15  to  fourths.  Ans.  -6¥0-. 

4.  Reduce  161T^  to  sixteenths. 

5.  Reduce  171||  to  an  improper  fraction.  Ans.  -H-V4— • 

6.  Change  1 1  to  a  fractional  form.  Ans.  *£■* 

7.  Change  100  to  an  improper  fraction. 

N   14 


158  COMMON   FRACTIONS. 

8.  Change  5  to  a  fraction  whose  denominator  shall  be  17. 

Ans.  f- f- . 

9.  Reduce  98§f  to  an  improper  fraction.  Ans.  -^yf--. 

10.  Reduce  116$f  to  an  improper  fraction.         Ans.  x£?z. 

11.  718||-  equal  how  many  ninety-sevenths  ?     Ans.  —g-6/-1. 

12.  Reduce  100i§J  to  an  improper  fraction.     Ans.  2£g.<p. 

13.  Reduce  7  to  an  improper  fraction. 

14.  Reduce  19  to  a  fraction  whose  denominator  shall  be  13. 

Ans.  *tf» 

15.  11G£  yards  equal  how  many  fourths  of  a  yard? 

Ans.  465  fourths. 

223.  To  reduce  a  compound  fraction  to  a  simple  fraction. 
Ex.  1.     Reduce  f  of  £  to  a  simple  fraction.  Ans.  §£. 

By  multiplying  the  denominator  of  |  by 

operation.  4^  tlie^  denominator  of  f ,  it  is  evident,  we 

i  X  $  =  "si  >  Ans.     obtain  \  of  \  =  572,  since  the  parts  into 

which  the  number  is  divided  are  4  times 

as  many,  and  consequently  only  i  as  large  as  before  ;  and  since  I  of 

1  =  32 '  i  of  s  wiU  De  3  times  53  =  \\  • 

Rule.  —  Multiply  all  the  numerator*  tor/ether  for  a  new  numerator, 
and  all  the  denominators  for  a  new  denominator. 

Note  1.  —  All  whole  and  mixed  numbers  in  the  compound  fraction  must  be 
reduced  to  improper  fractions,  before  multiplying. 

Note  2.  —  When  there  are  factors  common  to  both  numerator  and  denomi- 
nator, they  may  be  cancelled  in  the  operation. 

Examples. 

2.  Reduce  %  of  y\  of  £|  of  § §  to  a  simple  fraction.   Ans.  ^4 

OPERATION. 

1 

$      *      .re     $0 

^x^xg0X3l==TT'Ans- 

11 

3.  What  is  §  of  I  of  I  of  \i  ?  Ans.  tffr  =  f  f  f . 

4.  What  is  f-  of  ft  of  f  of  TV  ? 

5.  Reduce  y  of  f  of  f  of  |J  to  a  simple  fraction. 

6.  What  is  the  value  of  T6T  of  f  of  i  of  21  ? 

Ans.  HI  -  2TV 


COMMON   FRACTIONS.  159 

7.  What  is  the  value  of  T7T  of  15J  of  5^  of  100  ? 

Ans.  5758JJ . 

8.  What  is  £  off  of£J? 

9.  What  is  the  value  of  T7T  of  £$  of  |f-  of  $  7|  ? 

Ans.  $1.75. 

10.  What  is  the  value  of  £  of  T9r  of  ■} -J  of  3f  gallons  ? 

Ans.  |  gal. 

11.  What  part  of  a  ship  is  ^  of  §  of  f  ? 

12.  What  is  the  value  of  }  of  &  of  ||  of  |f  of  $  34  ? 

Ans.  $  6.75. 

A   COMMON   DENOMINATOR 

224  #  Fractions  have  a  common  denominator  when  all  their 
denominators  are  alike. 

225 1  A  common  denominator  of  two  or  more  fractions  is  a 
common  multiple  of  their  denominators ;  and  their  least  com- 
mon denominator  is  the  least  common  multiple  of  their  denomi- 
nators. 

226.     To  reduce  fractions  to  a  common  denominator. 

Ex.  1.  Eeduce  |-,  Tj,  and  |£  to  other  fractions  of  equal 
value,  having  a  common  denominator. 

FIRST     OPERATION. 

7X12X16  =  1344  new  numerator.    £    =  jf  || } 
5X     8X16=     640     "  "  A=VWfe[-Ans. 

11X8X12  =  1056"  "         a  =  iet ) 


8X12X16  =  153  6  common  denominator. 

We  first  multiply  the  numerator  of  J  by  the  denominators  1 2  and 
16,  and  obtain  1344  for  a  new  numerator.  We  next  multiply  the 
numerator  of  a  by  the  denominators  8  and  16,  and  obtain  640  for  a 
new  numerator ;  and  then  we  multiply  the  numerator  of  {\  by  the  de- 
nominators 8  and  12,  and  obtain  1056  for  a  new  numerator.  Finally, 
we  multiply  all  the  denominators  together  for  a  common  denominator, 
and  write  it  under  the  several  numerators,  as  in  the  operation. 

By  this  process,  since  the  numerator  and  denominator  of  each  frac- 
tion are  multiplied  by  the  same  numbers,  their  relation  to  each  other 
is  not  changed,  and  the  value  of  the  fraction  remains  the  same. 
(Art.  217.) 


12 

|16 

8 

~~3 

12 

16 

160  COMMON   FRACTIONS. 


SECOND    OPERATION. 

48    least  common  denominator. 

6  X    7  =  42,  new  numerator.  £  =  || } 
4X    5  =  20,    "  «  TV  =  f$f-Ans. 

3XH-33,    "  «  &  =  «) 

16  X  3  =  48,  least  common  multiple,  and  least  common  denomina- 
tor. 

Having  first  obtained  the  least  common  multiple  of  all  the  denomi- 
nators of  the  given  fractions,  we  assume  this  to  be  their  least  common 
denominator.  We  then  take  such  a  part  of  thi*  number,  48,  as  is  ex- 
pressed by  each  of  the  fractions  separately  for  their  respective  new 
numerators.  Thus,  to  get  a  new  numerator  for  -J,  we  take  -J  of  48, 
the  least  common  denominator,  by  dividing  it  by  8,  and  multiplying  the 
quotient  6  by  7.  We  proceed  in  like  manner  with  each  of  the  frac- 
tions, and  write  the  numerators  thus  obtained  over  the  least  common 
denominator.  In  this  process  the  value  of  each  fraction  remains 
unchanged,  as  both  terms  are  multiplied  by  the  same  number. 
(Art.  217.) 

The  method  used  in  the  second  operation,  it  will  be  perceived,  ex- 
presses the  fractions  of  the  result  in  lower  terms  than  that  used  in  the 
first.     On  this  account  it  is  often  to  be  preferred  to  the  other. 

Rule.  —  Find  the  least  common  multiple  of  the  denominators  for  the 
LEAST  COMMON  denominator. 

Divide  the  least  common  <1<  nominator  by  the  denominator  of  each  of 
the  given  fractions,  and  multiply  the  Quotients  by  their  respective  numer- 
ators, for  the  new  numerators.     Or, 

Multiply  each  numerator  by  all  the  denominators  except  its  aim,  for 
the  new  numerators  ;  and  all  the  denominators  together  for  A  common 
denominator. 

Note  1.  —  Compound  fractions  must  be  reduced  to  simple  ones,  whole  and 
mixed  numbers  to  improper  fractions,  before  finding  a  common  denominator, 
and  all  to  their  lowest  terms,  before  finding  the  least  common  denominator. 

Note  2.  —  Fractions  may  sometimes  be  reduced  to  a  common  denominator 
most  readily  by  multiplying  both  terms  of  one  or  more  of  them  by  such  a 
number  as  will  make  all  the  denominators  alike.  Thus  £  and  £  may  be 
brought  to  a  common  denominator  simply  by  multiplying  both  terms  of  the  £ 
by  2 ,  and  changing  in  that  way  its  form  to  f . 

Note  3.  —  Fractions  may  often  be  reduced  to  lower  terms,  without  destroy- 
ing their  common  denominator,  by  dividing  all  their  numerators  and  denomi- 
nators by  a  common  divisor. 

Examples. 
Reduce  the  following  fractions  to  their  least  common  denom- 
inator :  — 

2.  Reduce  f ,  f ,  J,  and  ft.  Ans.  ft  ft  ft  ft 

3.  Reduce  T\,T\,ig,  and  A.    Ans.  &&,  &&,  *§ft  &&. 


COMMON  FEACTIONS, 


161 


4. 

5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 


#,  T*T,  * ,  and  f . 


Reduce  |,  y3^  /T,  and  ■£?. 
Reduce  Jf ,  f ,  j£,  and  J. 
Reduce 
Reduce 
Reduce  f- ,  f ,  T5T,  and  T7T. 
Reduce  f- ,  J,  ^,  and  3f . 
Reduce  T52,  f,  f,  and  4|. 
Reduce  f,  T5¥,  ^f,  and  £. 
Reduce  f ,  -/^  -J- J,  and  /T. 
Reduce  £&  ^-,  £J,  and  ^. 
Reduce  J,  «^r,  T7F,  and  /^j. 
Reduce  § ,  7,  8,  and  5f . 


i  f ,  |,  and 


Ans. 


Arm  32  24  33  21 
-^S^  ¥2?  ¥2?  ¥2-?  ¥2* 
210   84   176  231 

ttd'fj  "3"  0  8"?  inrg">  uaB- 

30  40  45  48 
^<T>  "50-»  ^0~>  SJJ' 
An«  594  308   3  15  441 
^•ns»  -6"FTJJ  "6"93")  ^F3">  "6"9TT* 


Ans. 


Ans, 


798 
T"6"8- 


70    63 
TFF?  T6B? 

2  4   15       21< 


Ans.  f  J,  i| 


>  ¥2'  ¥; 


Ane  224  147  238   24 
^nS-  252?  252?  2  525  2.52* 
An«  570  140  165   12 

^Lns.  -g-uQ-,  -g-^Q-,  -go-^j,  -ggxr' 

And  210  140  105   84 
AUS.  2-yg-J  2£3">  2¥0">  2¥(T 


Reduce  the  following  fractions  to  a  common  denominator :  — 


tV 


16.  Reduce  f,  f,  and  ' 
nominator. 

17.  Reduce  f ,  f ,  and  T%- 

18.  Reduce  T6T,  f ,  and 

19.  Reduce  TV,  T%  and  7f . 

20.  Reduce  -ff,  f,  and  TV 

21.  Reduce  f,  T4T,  and  11TV 

22.  Reduce  £,  §,  £,  and  8. 

23.  Reduce  f,  T7T,  and  f  of  7|. 
Reduce  ^,  f ,  £,  and  17. 
Reduce  $J,  £  of  6,  and  21  J. 


to  fractions  having  a  common  de- 

A^q       80         96        105 
Allb.  J2(T)  T2<T>  T2TT' 

Ans. 
Ans. 


360      560      189 
USO"?  -BSU)  ^UTT- 
_5  4  6_    _5  7  2      _6  1  6 
10"0~1>    10~0T>  TO~0~T' 


Ant-      1485      1020       612 
■AnS«    2  2  9"5J  22F"5>  22  ?? 

Ans 

Ans 

Anq   528    7  56   6039 
AnS*  TTF8-?  TTF8"?  TTBB" 


204  0  _5_4  0_ 
22F5?  22F5? 
21  28  24 

¥2?  ¥2?  ¥2? 


263  16 
"2  2 ¥  5"' 


24. 
25. 

26. 


Ans. 


110 
T2U> 


5.X6    2- 
T20?    T 


w- 


Reduce  f,  T4T,  T%  f ,  and  J. 


Anc     12012        5096         5390         8008         7007 
Ans.  T¥IFT¥>  T4<IT¥>  T¥tfT¥>  T¥T5T¥?  T¥ITT¥- 

27.  Reduce  T%,  $f$,  and  v^. 

An«   28506816    37088064     722813 
Ans.  TT¥4^T5  5  2?  TTF¥U15  5  2?  TtW^T552' 

ADDITION  OF   COMMON  FRACTIONS. 

227.     Addition  of  fractions  is  the  process  of  finding  the  value 
of  two  or  more  fractions  in  one  sum. 


Note.  —  Only  units  of  the  same  kind,  whether  integral  or  fractional,  can  be 
collected  into  one  sum ;  if,  therefore,  the  fractions  to  be  added  do  not  express 
the  same  fractional  unit,  they  require  to  be  brought  to  the  same,  by  being  re- 
duced to  a  common  or  the  least  common  denominator. 

228.     To  add  together  two  or  more  fractions. 
14* 


162  COMMON  FRACTIONS. 

Ex.  1.  Add  T32-,  T52-,  T72-,  and  {§  together.     Ans.  f  f  =  2£. 

operation.  These  fractions 

A  +  A  +  tV  +  U  =  f  I  =  ¥  =  H-     all  being  te*#fc, 

that  is,  having  12 
for  a  common  denominator,  we  add  their  numerators  together,  and 
write  their  sum,  26,  over  the  common  denominator,  12.  Thus  we 
obtain  |-|,  which,  being  reduced,  =  2^,  the  sum  required. 

2.  What  is  the  sum  of  },  fo  \&  and  £g  ?  Ans.  2$j$. 

$     12     16  120 


3        4  8 

12 

20  X  4  X  3  =  240   16 

20 


OPERATION. 

2  4  0  =  least  common  denominator. 


30  X      7  =  210 

20X      5  =  100  I  new  numera- 

15X11  =  165  f         tors. 
12x13  =  156 


Sum  of  numerators,  6  3  1 

' 2151   Ans. 

Least  com.  denom.,  2  4  0  2?TT'  ' 

The  given  fractions  not  expressing  the  same  kind  of  fractional 
unit,  we  reduce  them  to  their  least  common  denominator,  and  thus 
make  the  fractional  parts  all  of  the  same  kind.  The  fractions  now 
all  expressing  two-hundrcd-fortieths,  we  add  their  numerators,  and 
write  the  result,  631,  over  the  least  common  denominator,  240,  and 
obtain  f|^  =  2^fJ,  the  answer  required. 

Rule.  —  Reduce  the  fractions,  if  necessary,  to  a  common,  or  the  least 
common  denominator,  and  write  the  sum  of  the  numerators  over  their 
common  di  nominator. 

Note  1.  —  Mixed  numbers  must  be  reduced  to  improper  fractions,  and  com- 
pound fractions  to  simple  fractions,  and  each  fraction  to  its  lowest  terms,  be- 
fore attempting  to  obtain  the  common  denominator. 

Note  2.  —  In  adding  mixed  numbers,  the  fractional  parts  may  be  added 
separately,  and  their  sum  added  to  the  amount  of  the  whole  numbers. 

Examples. 

3.  Add  T5T,  T6T,  T9T,  \\,  -J4,  and  |f  together.      Ans.  3-ff 
,    4.  Add  ,fe  &,  H,  i%,  and  £f  together.  Ans.  2*J. 

5.  What  is  the  sum  of  ^T,  ^-f ,  |f ,  and  £f  ? 

6.  What  is  the  sum  of  TW,  T\%  T\%  and  |ff  ?     Ans.  If. 

7.  What  is  the  sum  of  |Jf,  |§ j,  |f {,  and  fa  ? 

Ans.  mi. 

8.  Add  J,  £, 1£,  and  f  together. 

9.  Add  -JT,  /g-,  -J,  and  \  together.  Ans.  lffj. 
10.  Add  T\,  y5F>  f$,  and  J  together.  Ans.  2£f£. 


COMMON  FRACTIONS.  163 

11.  Add  T5T,  &>  rtfe  an(*  i  together.  Ans.  1. 

12.  Add  -f-f ,  J-J,  £&  and  -&  together.  Ans.  3£9. 

13.  Add  },  i,  &  I,  &  and  j-  together.  Ans.  1T\\. 

14.  Add  f ,  $,  and  5f  together.  Ans.  6f§f. 

15.  Add  -^T,  /^  and  9T3T  together. 

16.  Add  f,  |,  and  4f  together.  Ans.  6  J. 

17.  Add  |,  7£,  and  8f  together.  Ans.  17^V 

18.  Add  £,  3£,  and  5f  together.  Ans.  9^J. 

19.  Add  6f,  7f,  and  4f  together.  Ans.  18£^. 

20.  What  is  the  sum  of  17$,  14£,  and  13  f  ? 

21.  What  is  the  sum  of  16§,  8£,  9f,  3£,  and  1  J-  ? 

Ans.  40TV 

22.  What  is  the  sum  of  371-J-J,  614 Jg-,  and  81f  ? 

Ans.  1068fJ. 

23.  Add  f  of  18T3T,  and  ■&  of  f  of  6T3T  together. 

Ans.  12ff£. 

24.  Add  £  of  18,  and  T4T  of  -fJ-  of  7T\  together. 

229.     To  add  any  two  fractions,  whose  numerators  are  alike. 

Ex.  1.  Add  I  to  l>  Ans.  ^. 

operation.  \Ye  first  find  the 

Sum  of  the  denominators,         5  -}-  4  =    9       sum  of  the  denom- 

Product  of  the  denominators,  4x5  =  20     i^tor^   ™hic}\  .is 

9,  and  then  their 

product,  which  is  20 ;  and  the  9  being  written  as  a  numerator  of  a 

fraction,  and  the  20  as  its  denominator,  the  result,  -^,  is  the  answer 

required.     The  reason  of  the  operation  is,  that  the  process  reduces 

the  fractions  to  a  common  denominator,  and  then  adds  their  numerators. 

Hence,  to  add  two  fractions  whose  numerators  are  a  unit, 

Write  the  sum  of  the  given  denominators  over  their  product. 

2.  Add  £•  to  f  Ans.  lfa 

Sum  of  the  denominators  X 


OPERATION. 


by  one  of  the  numerators,  (4  +  5)  X  3        27 

"  _ ! '_ 1    7        AriS 

Product  of  the  denominators,        4x5  "20 

By  multiplying  the  sum  of  the  denominators  by  one  of  the  numer- 
ators for  a  new  numerator,  and  the  denominators  together  for  a  new 
denominator,  we  reduce  the  fractions  to  a  common  denominator,  and 
add  their  numerators,  and  thus  obtain  -|J  =  l^g-,  the  answer  re- 
quired. Hence,  to  add  fractions  whose  numerators  are  alike,  and 
greater  than  a  unit, 

Write  the  product  of  the  sum  of  the  given  denominators  by  one  of  the 
numerators  over  the  product  of  the  denominators. 


164  COMMON   FRACTIONS. 

Examples. 

3.  Add  &  to  i,  i  to  i  i  to  £,  i  to*,  *  to  i,  $  to  jfc  i  to  f 

4.  Add  TV  to  J,  TV  to  i,  T\  to  J,  JT  to  |,  JT  to  i,  T\  to  f 
o.  Add  TV  to  i,  TV  to  i  TV  to  |,  TV  to  |,  TV  to  J,  TV  to  f 

6.  Add  J  to  J,  J-  to  £,  i  to  i,  J  to  i,  i  to  fc  i  to  |,  J-  to  J. 

7.  Add  *  to  $,  |  to  J,  i  to  i,  |  to  £,  i  to  i,  i  to  i,  i  to  J. 

8.  Add  }  to  *,  |  to  J,  |  to  i,  |  to  i,  |  to  i,  |  to  4,  -f  to  £. 

9.  Add  iU>i,i  to  J,  4  to  i,  i  to  *,  4  to  &  i  to  |,  i  to  f 

10.  Add  f  to  T8T,  f  to  &,  f  to  A,  f  to  T«V,  f  to  T»^,  f  to  jft-. 

11.  Add  ft  to  *,  |  to  f,  |  to  TS  f  to?,  f  tof,  f  to  T2T,  f  to  f 

12.  Add  §  to  -ft-,  f  to  f,  &  to  f  £  to  f,  £  to  £,  £  to  f>T. 

13.  Add  f  to  f,  f  tO  -ft-,  }  tO  ^  T6T  tO  T%  T6T  tO  f7,  T6T  to  T%. 

u.  Add  f  to  T«T,  tbt  to  -ft.,  a  to  T^,  a  to  ^  A  to  jy,  a  to  Tv 

15.  Add  A  to  T9T,  A  to  T9¥,  A  to  Ai  A  to  T%  A  to  A- 

SUBTRACTION  OF  COMMON  FRACTIONS. 

230.     Subtraction  of  Fractions  is  the  process  of  finding 

the  difference  between  two  fractions. 

Note.  —  When  the  fractions  express  different  fractional  units,  they  require 
to  be  brought  to  those  of  the  same  kind  before  the  subtraction  can  be  per- 
formed.   • 

To  subtract  one  fraction  from  another. 

Ex.  1.  From  \$  take  A-  Ans-  t?  —  h 

operation.  The  fractions  both  being  twelfths,  having  12 

jj-  —  A  =  tV     f°r  a  common  denominator,  we  subtract  the 

less  numerator  from  the  greater,  and  write  the 

difference,  6,  over  the  common  denominator,  12.     Thus,  we  have 

A  as  the  difference  required. 

2.  From  tf  take  f  Ans.  f  |  =  ££ . 

operation.  The  given  frac- 

7  7  common  denominator.  tions  not  express- 

-i  -i  [    7  v  -i  s\ n  r\  \  ing  the  same  kind 

7iiv      A-.A  At  new  numerators.  of  fractional  unit, 

/ 1 1  i  X      *  —  44  )  we   reduce  them 

2  6      dif.  of  numerators.  to  a  common  de- 

— -  .  nominator,      and 

4  8      common  denominator,    thus      make    the 

fractional  parts  all 
of  the  same  kind.  We  next  find  the  difference  of  the  new  numera- 
tors, which  we  write  over  the  common  denominator,  and  obtain 
ff ,  the  answer  required. 


COMMON   FRACTIONS.  165 

Rule.  — Reduce  the  fractions,  if  necessary,  to  a  common,  or  the  least 
common  denominator.  Write  the  difference  of  the  numerators  over  their 
common  denominator. 

Note.  —  If  the  minuend  or  subtrahend,  or  both,  are  compound  fractions, 
they  must  be  reduced  to  simple  ones. 

Examples. 

3.  Subtract  T6T  from  ±f.  Ans.  T7T. 

4.  Subtract  T4F  from  -£§ .  Ans.  T9F. 

5.  From  §f  take  ^8T. 

6.  From  £f  take  £$.  Ans.  &. 

7.  From  f  f  take  £$.  Ans.  ^. 

Ans.  £. 


8.  From  §f  take  £f 

9.  Subtract  ^-f-  from  f  J.  jsois.  £. 

10.  Subtract  T\\  from  T3¥6¥.  Ans.  -fa. 

11.  Subtract  T^  from  ^-fa.  Ans.  ^ 

12.  Subtract  T3T2/F  from  T5T9/B.  Ans.  T2T%. 

13.  Subtract  T£§ ^  from  TV°xyV  Ans.  W^. 

14.  From  \\  take  ^. 

15.  From  -/^  take  x5^.  Ans.  ^J. 

16.  From  \\  take  TV  Ans.  §f£. 

17.  From  T|  take  &.  Ans.  f |f. 

18.  From  ^  take  T4T.  Ans.  J^T. 

19.  From  f£  take  T%. 

20.  From  ^z  take  ^  Ans.  y3^. 

21.  From  £$  take  T5g.  Ans.  &, 

22.  From  ^  take  ^.  Ans.  -fe. 

23.  From  7f  take  f  of  9.  Ans.  l^. 

24.  What  is  the  value  of  f  of  8f  —  §  of  5  ? 

25.  What  is  the  value  of  i  of  3  —  £  of  2  ?  Ans.  jfc 

281.     To  subtract  a  proper  fraction  or  a  mixed  number 
from  a  whole  number. 

Ex.  1.  From  7  take  3f.  Ans.  3f. 

operation.  Since  we  have  no  fraction  from  which  to  subtract 

From  7        *ne  1 »  we  must  add  1,  or  its  equal,  -§,  to  the  minuend, 
Take   3  6      an(^  sa^  "§"  ^rom  I"  ^eaves  f  •     We  WI"ite  the  |  below 
*     the  line,  and  carry  1  to  the  3  in  the  subtrahend,  and 
Rem.  3 §-      subtract  as  in  subtraction  of  simple  whole  numbers. 
The  result  will  be  obtained,  if  we 


166  COMMON   FRACTIONS. 

» 
Subtract  the  numerator  from  the  denominator  of  the  fraction,  and 
under  the  remainder  write  the  denominator,  and  adding  one  to  the  ichole 
number  in  the  subtrahend,  subtract  the  sum  from  the  minuend. 

Note.  —  When  the  subtrahend  is  a  mixed  number,  we  may  reduce  it  to  an 
improper  fraction,  and  change  the  whole  number  in  the  minuend  to  a  fraction 
having  the  same  denominator,  and  then  proceed  as  in  Art.  230. 


Examples. 

2. 

3. 

4. 

5. 

6. 

7. 

From  3  2 

16 

671 

385 

1  6 

18 

Take      5| 

H 

o» 

im 

om 

If 

Eem.  2  6£ 

ll* 

6  7  0& 

3  68ff 

i*m 

16* 

8. 

9. 

10. 

11. 

12. 

13. 

From  1  9 

27 

169 

711 

46 

81 

Take  1  3f 

** 

&1** 

8  0ff 

15f 

4  9W 

FIRST 

OPERATION. 

From  8f  = 

-m 

Take 

H- 

-   4.28 

232.     To  subtract  one  mixed  number  from  another  mixed 

number. 

Ex.  1.  From  8?  take  4|.  Ans.  3ff. 

We  first  reduce  the  fractional  parts  to  a  com- 
mon denominator,  and  obtain  as  their  equivalents 
4-|  and  |4.  Now,  since  we  cannot  take  -f  |  from 
-||,  we  add  1,  equal  to  |4,  to  the  \%  in  the  min- 

Rem.  3ff    uend,  and  obtain  f£.     From  f£  taking  ff ,  we 

have  left  $-§,  which  we  write  below  the  line,  and 

carry  1  to  the  4  in  the  subtrahend,  and  subtract  from  the  8  above  as 

in  subtraction  of  simple  whole  numbers. 

second  operation.  In  this  operation,  we  reduce 

From  8f  =  -^  =  2$£-  the  mixed  numbers  to  improper 

Take  4^  =  2Jt-  =  -'t^-8-  fractions,  and  these  fractions  to 
a  common    denominator.      We 

Rem.  *f£-  =  3§f    then  subtract  the  less  fraction 

from  the  greater,  and,  reducing 

the  remainder  to  a  mixed  number,  obtain  3||-,  as  before.     Hence,  in 

performing  like  examples,  we  may 

Reduce  the  fractional  parts,  if  necessary,  to  a  common  denominator, 
and  subtract  the  fractional  parts  and  the  ichole  numbers  separately. 
Increase  the  fractional  part  of  the  minuend,  when  otherwise  it  would  be 
less  than  the  subtrahend,  before  subtracting,  by  as  many  parts  as  it  takes 
to  make  a  unit  of  the  fraction  (Art.  208),  and  carry  1  to  the  whole 
number  of  the  subtrahend  before  subtracting  it.     Or, 

Reduce  the  mixed  numbers  to  improper  fractions,  then  to  a  common 
denominator,  and  subtract  the  less  fraction  from  the  greater. 


CL 


COMMON   FRACTIONS. 


167 


Examples. 


2. 

3. 

4. 

5. 

6. 

7. 

cwt. 

Tuns. 

1 

lb. 

OZ. 

Miles. 

From  1  8| 

7  3| 

6  7* 

2  9lf 

144£ 

171M 

Take      9f 

16T2" 

16| 
10. 

15| 

9  9|x 

»1* 

8. 

9. 

11. 

12. 

13. 

Furlongs. 

Rods. 

Inches. 

Feet. 

Bushels. 

Pecks. 

From  1  0  1£ 

165^ 

*7« 

8  4f| 

6  7Hii 

17«? 

Take      9  3| 

9  8J 

l»i» 

1«« 

18  3|1 

«m 

14.  From  a  hogshead  of  wine  there  leaked  out  7T9T  gallons ; 
what  quantity  remained  ?  Ans.  55T^-gal. 

15.  A  man  engaged  to  labor  30  days,  but  was  absent  5^ 
days  ;  how  many  days  did  he  work  ? 

16.  From  144  pounds  of  sugar  there  were  taken  at  one 
time  17f  pounds,  and  at  another  28^  pounds ;  what  quantity 
remains  ?  Ans.  97^-f  lb. 

17.  A  man  sells  9  J-  yards  from  a  piece  of  cloth  containing 
34  yards  ;  how  many  yards  remain  ?  Ans.  24|yd. 

18.  The  distance  from  Boston  to  Providence  is  40  miles. 
A,  having  set  out  from  Boston,  has  travelled  -fY  of  the  distance ; 
and  B,  having  set  out  at  the  same  time  from  Providence,  has 
gone  T^-  of  the  distance ;  how  far  is  A  from  B  ? 

Ans.  28T|Tm. 

19.  From  \  of  a  square  yard  take  ^  of  a  yard  square. 

233.  To  subtract  one  fraction  from  another,  when  their 
numerators  are  alike. 


Ex.  1.  From  £  take  |. 


Ans. 


A- 


OPERATION. 


7  —  3  =  4,    difference  of  the  denominators. 
7  X  3  =  21,  product  of  the  denominators. 

We  first  find  the  product  of  the  denominators,  which  is  21,  and 
then  their  difference,  which  is  4,  and  write  the  former  for  the  denom- 
inator of  the  required  fraction,  and  the  latter  for  the  numerator. 
By  this  process  the  fractions  are  reduced  to  a  common  denominator, 
and  their  difference  found.  Hence,  to  subtract  one  fraction  from 
another,  whose  numerators  are  a  unit,  we  may 

Write  the  difference  of  the  denominators  over  their  product. 


168  COMMON   FEACTIONS. 

2.  Take  f  from  f .  Ans.  &. 


OPERATION. 


'Ans. 


Difference  of  the  denominators  mul- 
tiplied by  one  of  the  numerators,     (7  —  3)  X  2 8 

Product  of  the  derominators,  3x7  21 

We  multiply  the  difference  of  the  denominators  by  one  of  the 
numerators  for  a  new  numerator,  and  the  denominators  together 
for  a  new  denominator,  by  which  process  the  fractions  are  reduced 
to  a  common  denominator,  and  the  difference  of  their  numerators  is 
found.  Hence,  when  the  given  fractions  have  their  numerators  alike 
and  greater  than  a  unit,  we  may 

Write  the  product  of  the  difference  of  the  given  denominators,  by  one 
of  the  numerators,  over  the  product  of  the  denominators. 


17.  Take  T8T  from  f- ;  T8T  from  f  ;  T8T  from  | ; 

18.  Take  T2T  from  f  ;  T2T  from  f  ;  T2T  from  §  ; 

19.  Take  ft  from  Xp- ;  ft  from  ft ;  ft  from  ft ;  ft  from 

20.  Take  \2-  from  ■£ ;  -V2-  from  *£ ;  -1/  from  ^ ;  ft  from 

MISCELLANEOUS  EXAMPLES  IN  ADDITION  AND   SUB- 
TRACTION OF  ERACTIONS. 

1.  A  benevolent  man  has  given  to  one  poor  family  J  of  a 
cord  of  wood,  to  another  £  of  a  cord,  and  to  a  third  ^  of  a  cord ; 
how  much  has  he  given  to  them  all  ?  Ans.  2£  cords. 


COMMON   FRACTIONS.  169 

2.  I  have  paid  for  a  knife  $  f ,  for  a  Common  School  Arith- 
metic $  £,  for  a  slate  $  -£,  and  for  stationery  $  J ;  what  did  I 
pay  for  the  whole  ? 

3.  R.  Howland  travelled  one  day  20T70-  miles,  another  day 
19  J-  miles,  and  a  third  day  22T16-  miles ;  what  was  the  whole 
distance  travelled  ?  Ans.  62§£  miles. 

4.  I  have  bought  6 J  tons  of  anthracite  coal,  19|-  tons  of 
Cumberland  coal,  and  3  j-  tons  of  cannel  coal ;  what  is  the  whole 
quantity  purchased  ?  Ans.  30^  tons. 

5.  There  is  a  pole  standing  £  in  the  mud,  ^  in  the  water,  and 
the  remainder  above  the  water ;  what  portion  of  it  is  above  the 
water  ? 

6.  F.  Adams,  having  a  lot  of  sheep,  sold  at  one  time  §  of 
them,  and  at  another  time  £  of  the  remainder ;  what  portion 
of  the  original  number  had  he  then  left  ?  Ans.  ^. 

7.  From  a  piece  of  calico  containing  31£  yards  there  have 
been  sold  llf  yards,  9£  yards,  and  3f-  yards;  how  much  re- 
mains ? 

8.  From  a  cask  of  molasses  containing  84|  gallons,  there 
were  drawn  at  one  time  4f-  gallons,  at  another  time  11  gal- 
lons ;  at  a  third  time  2  6 J-  gallons  were  drawn,  and  £  of  7J 
gallons  returned  to  the  cask ;  and  at  a  fourth  time  13T8T  gallons 
were  drawn,  and  3  J-  gallons  of  it  returned  to  the  cask.  How 
much  then  remained  in  the  cask?  Ans.  35  Jf- Jgal. 

9.  A  merchant  had  3  pieces  of  cloth,  containing,  respectively, 
19 j-  yards,  36^-  yards,  and  33 1  yards.  After  selling  several 
yards  from  each  piece,  he  found  he  had  left  in  the  aggregate 
71  f  yards.     How  many  yards  had  he  sold  ?  Ans.  18^. 

MULTIPLICATION  OF   COMMON  FRACTIONS. 

234 1  Multiplication  of  Fractions  is  the  process  of  multi- 
plying when  the  multiplier,  or  multiplicand,  or  both,  are  frac- 
tional numbers. 

Note.  —  If  the  multiplier  is  less  than  1,  only  such  a  part  of  the  multiplicand 
is  taken  as  the  multiplier  is  of  1.  Therefore,  the  product  resulting  from 
multiplying  a  number  by  a  proper  fraction  is  not  larger,  but  less,  than  the 
multiplicand. 

235.  To  multiply  when  one  or  both  of  the  factors  are  frac- 
tions. 

N    15 


170  COMMON   FRACTIONS. 

Ex.  1.     Multiply  ^  by  9.  Ans.  ff  =  3*. 

first  operation.  ft  is  evident  that  the  fraction 

T3"  X  9  ==  f|  =  J  =  H  Ans.      iV  is  multiplied  by  9  by  multi- 
plying its  numerator  by  9,  since 
the  parts  taken,  63,  are  9  times  as  many  as  before,  while  the  parts 
into  which  the  unit  of  the  fraction  is  divided  remain  the  same. 

second  operation.  ft  is  evident,  also,  that  the  fraction 

tV  X  0  =  %  =  3£  Ans.     X  is  multiplied  by  9  by  dividing  its 

denominator  by  9,  since  the  parts  into 
which  the  unit  of  the  fraction  is  divided  are  only  \  as  many,  and  con- 
sequently 9  times  as  large,  as  before,  while  the  parts  taken  remain 
the  same.     Therefore, 

Multiplying  the  numerator  or  dividing  the  denominator  of  a  fraction 
by  any  number  multiplies  the  fraction  by  that  number  (Art.  217). 

2.  Multiply  14  by  f  Ans.  G. 

first  operation.  By  dividing  the  whole  number,  14,  by 

7)14  7>  the  denominator  of  the  fraction,  we 

obtain  ^  of  14  =  2,  which  multiplied  by 

2X3=6  Ans.     3,  the  numerator  of  the  fraction,  gives  $ 

of  14  =  6. 

second  operation.  By  multiplying  the  whole  number,  14, 

14  by  3,  the  numerator  of  the  fraction,  we 

3  obtain  42,  a  product  7  times  as  large  as 

~rr        „  .  it  should  be,  as  the  multiplier  was  not  3, 

4  ^  "*"  '  ==  "  Ans.  a  ^ota  number,  but  -| ,  or  3  -f-  7  ;  hence, 

we  divide  the  42  by  7 ;  and  thus  obtain 
^  of  14  =  6,  as  before.     Therefore, 

Multiplying  by  a  fraction  is  taking  the  part  of  the  multiplicand  de- 
noted by  the  multiplu  r. 

3.  Multiply  i  by  f .  Ans.  f  J. 

operation.  To  multiply  -J  by  f  is  to  take  f  of  the 

J  X  -|  =  f  tj  Ans.     multiplicand,  -|.     Now,  to  obtain  -|  of  -|,  we 

multiply  the  numerators  together  for  a  new 

numerator,  and  the  denominators  together  for  a  new  denominator 

(Art.  226).     Therefore, 

Multiplying  one  fraction  by  another  is  the  same  as  reducing  com- 
pound fractions  to  simple  ones. 

When  either  of  the  factors  is  not  a  fraction,  as  in  examples  first  and 
second,  it  may  be  reduced  to  a  fractional  form,  and  then  the  operation 
may  be  like  that  in  the  last  example.     Hence  the  general 

Rule.  —  Reduce  whole  or  mixed  numbers,  if  any,  to  improper  frac- 
tion*. Multiply  the  numerators  together  for  a  new  numerator,  and  the 
denominators  together  for  a  new  denominator. 

Note.  —  "When  there  are  common  factors  in  the  numerators  and  denomina-; 
tors,  the  operation  may  be  shortened  by  cancelling  those  factors. 


COMMON  FRACTIONS.  171 


Examples. 

N 

4.  Multiply  T7T  by  ft 

Ans.  -J-. 

OPERATION. 

jr  a    l 

s/ —  —  • 

11     %t      3 
3 

5.  Multiply  12  by  f            x 

Ans.  8f 

6.  Multiply  f  by  12. 

7.  Multiply  \\  by  J$. 

Ans.  £. 

8.  Multiply  f  by  |£ 

Ans.  yW 

9.  Multiply  f  by  T6T. 

Ans.  ||. 

10.  Multiply  A-  by  if 

Ans.  T79V 

11.  Multiply  T8T  by  f, 

Ans.  f  f 

12.  Multiply  f  by  -if. 

13.  Multiply  if  by  |J. 

Ans.  |f 

14.  Multiply  T9^  by  14. 

Ans.  12f. 

15.  Multiply  13  by  f. 

Ans.  7f 

16.  Multiply  16  by  TV 

Ans.  2T§. 

17.  Multiply  11  by  f. 

Ans.  6f . 

18.  Multiply  ■&  by  14. 

19.  Multiply  |  by  19. 

Ans.  16f. 

20.  Multiply  TT  by  ff. 

Ans.  §. 

21.  Multiply  -|  by  if. 

Ans.  TV 

22.  Multiply  T\  by  § f . 

Ans.  -&. 

23.  Multiply  %  by  TT. 

24.  Multiply  ^  by  T^. 

Ans.  y^. 

25.  Multiply  $  of  T7T  of  H  by  100. 

Ans.  12£. 

26.  Multiply  i  of  f  of  £  by  11. 

Ans.  3ft. 

27.  What  cost  ^  of  a  ton  of  hay,  at  $  17 

per  ton  ? 

Ans.  $  9yi. 

28.  What  cost  ^  of  an  acre  of  land,  at  $  37  per  acre  ? 

29.  At  §  of  a  dollar  per  foot,  what  cost  7  cords  of  wood? 

30.  Multiply  leift  by  19££. 

Ans.  3136f3-. 

31.  Multiply  -f-  by  8f . 

Ans.  3f . 

32.  Multiply  T%  by  17T3T. 

Ans.  15TT. 

33.  Multiply  f  by  71|, 

Ans.  63I7;. 

34.  Multiply  J  of  9¥  by  §  of  17. 

Ans.  78 1. 

35.  Multiply  &  of  7  by  i|  of  87T3T. 

36.  Multiply  8  by  J. 

Ans.  6f . 

172  COMMON  FRACTIONS. 

37.  Multiply  12  by  f  Ans.  8f 

38.  Multiply  15  by  T6T.  Ans.  8T2T. 

39.  A  merchant  owning  £  of  a  ship  sells  T4T  of  his  share 
to  A.     What  part  is  that  of  the  whole  ship  ? 

40.  Multiply  3£  by  lOf  Ans.  39J-f. 

41.  Multiply  §  of  1\  by  \  of  llf.  Ans.  49|f|. 

42.  Multiply  f  of  9  by  f  of  17.  Ans.  26^. 

43.  Multiply  £  of  ST\  by  \  of  9f  Ans.  25?VV 

236.     When  one  of  the  factors  is  a  whole  number,  and  the 

other  a  mixed  number,  we  may 

Multiply  the  fractional  -part  and  the  whole  number  separately,  and 
add  together  the  products. 

Examples. 

I.  Multiply  7£-  by  9.  2.  Multiply  12  by  3£. 

OPERATION.  OPERATION. 

7*  12 

JL  -if 

|  X  9  =  V-=    5|  £ofl2=    9 

7x9  =63_  12X3  =  36 

68f  Ans.  45  Ans. 

3.  Multiply  8f  by  7.  Ans.  60£. 

4.  Multiply  17  by  3£. 

5.  Multiply  13  by  8f  Ans.  109f . 

6.  Multiply  37  by  13^.  Ans.  507|f. 

7.  Multiply  llf  by  8.  Ans.  94f 

8.  What  cost  7T6Tlb.  of  beef  at  5  cents  per  pound  ? 

9.  What  cost  23Ty)bl.  of  flour  at  $  6  per  barrel  ? 

Ans.  $141$. 
10.  What  cost  8|yd.  of  cloth  at  $  5  per  yard  ?    Ans.  $  41  J. 

II.  What  cost  9  barrels  of  vinegar  at  $  6f  per  barrel? 

Ans.  $  57f . 

12.  What  cost  12  cords  of  wood  at  %  6.37$  per  cord? 

Ans.  $7G.50. 

13.  What  cost  llcwt.  of  sugar  at  $  9f  per  cwt.  ? 

14.  What  cost  4|  bushels  of  rye  at  $  1.75  per  bushel  ? 

Ans.  $  7.G5f . 

15.  What  cost  7  tons  of  hay  at  $  11 -J-  per  ton  ? 

Ans.  $  83-|-. 


COMMON   FRACTIONS.  173 

16.  What  cost  9  dozen  of  adzes  at  $  10§  per  dozen? 

17.  What  cost  5  tons  of  timber  at  $3£  per  ton? 

Ans.  $15 1. 

18.  What  cost  15cwt.  of  rice  at  $7.62£  per  cwt.  ? 

Ans.  $114,371. 

19.  What  cost  40  tons  of  coal  at  $  8.37  J  per  ton  ? 

Ans.  $  335. 

DIVISION  OF   COMMON  FRACTIONS. 

237.  Division  of  Fractions  is  the  process  of  dividing  when 
the  divisor  or  dividend,  or  both,  are  fractional  numbers. 

Note.  —  If  the  divisor  is  less  than  1,  the  quotient  arising  from  the  division 
will  be  as  many  times  the  dividend  as  the  divisor  is  contained  times  in  1. 
Therefore,  the  quotient  arising  from  dividing  a  whole  or  mixed  number  by  a 
proper  fraction  will  always  be  larger  than  the  dividend. 

238.  The  reciprocal  of  a  fraction  is  the  number  resulting 
from  taking  its  numerator  as  denominator,  and  its  denominator 
as  numerator,  since  any  two  numbers,  whose  product  is  1,  are 
the  reciprocals  of  each  other.  Thus,  the  reciprocal  of  -^  is  that 
fraction  inverted,  or  t9q-,  since  JF°-  X  tit  =  *• 

239.  To  divide  when  the  divisor  or  dividend,  or  both,  are 
fractions. 

Ex.  1.  Divide  -ff  by  7.  Ans.  T2T. 

first  operation.  ft  is  evident  that  the  fraction  if  is  di- 

T f  -5-  7  =  T3jr  Ans.       vided  by  7  by  dividing  its  numerator  by  7, 

since  the  size  of  the  parts,  as  denoted  by 

the  denominator,  remains  the  same,  while  the  number  of  parts  taken 

is  only  T  as  large  as  before. 

second  operation.  ft  is  evident  the  fraction  is  also  di- 

^|.  -a-  7  =b  y1^-  =  T2T  Ans.     vided  by  7  by  multiplying  its  denomi- 
nator by  7,  since  the  number  of  parts 
taken,  as  denoted  by  the  numerator,  remains  the  same,  while  the 
size  of  the  parts  is  only  T  as  large  as  before.     Therefore, 

Dividing  the  numerator  or  multiplying  the  denominator  of  a  fraction 
by  any  number  divides  the  fraction  by  that  number  (Art.  217). 

2.  Divide  |f  by  T%.  Ans.  2. 

operation.  Since   tne  fractional  units  of  the  two 

t!  •*"  A  =  ^  Ans.     fractions    are    of  the    same    kind,    it,  is 
evident    that   12    thirteenths   contain    6 
15* 


174  COMMON   FRACTIONS. 

thirteenths  as  many  times  as  6  is  contained  in  12;  12  -r-  6  =  2,  Ans, 
Therefore, 

When  the  fractions  have  a  common  denominator,  the  division  can  be 
performed  as  in  whole  numbers,  by  dividing  the  numerator  of  the  divi- 
dend by  the  numerator  of  the  divisor, 

3.  Divide  }  by  f.  Ans.  ljf 

first  operation.  Having  reduced  the  frae- 

^-T-f  =  f|-7-fJ=  1^4"  Ans.     tions  to  a  common  denomi- 
nator,  we   divide    the   nu- 
merator 32  of  the  dividend  by  the  numerator  21  of  the  divisor,  as  in 
working  the  last  example,  and  obtain  as  the  required  result  1^\. 

second  operation.  In  the  second  oper- 

^-f-f  =  4-Xf  =  ff==  l^x  Ans.     ation,   we    invert   the 

divisor,  and  then  pro- 
ceed as  in  multiplication  of  fractions  (Art.  235).  The  reason  of  this 
process,  which  in  effect  reduces  the  fractions  to  a  common  denomi- 
nator, and  divides  the  numerator  of  the  dividend  by  that  of  the 
divisor,  will  be  seen,  if  we  consider  that  the  divisor,  -|,  is  an  expres- 
sion denoting  that  3  is  to  be  divided  by  8.  Now,  regarding  3  as  a 
whole  number,  we  divide  the  fraction  ^  by  it,  by  multiplying  the 
denominator;  thus,  4  x  8  =  -fc.  But  the  divisor  3  is  8  times  as  large 
as  it  ought  to  be,  since  it  was  to  be  divided  by  8,  as  seen  in  the  origi- 
nal fraction ;  then  the  quotient,  fa,  is  £  as  large  as  it  should  be,  and 
must  be  multiplied  by  8  ;  thus,  fa  *  8  =  £?  =  1%\,  the  answer,  as 
before.  By  this  operation  we  have  multiplied  the  dividend  by  the 
reciprocal  of  the  divisor,  the  denominator  of  the  dividend  having 
been  multiplied  by  the  numerator  of  the  divisor,  and  the  numerator 
of  the  dividend  by  the  denominator  of  the  divisor.     Therefore, 

Dividing  by  a  fraction  is  the  same  as  multiplying  by  its  reciprocal 

When  either  divisor  or  dividend  is  not  a  fraction,  it  may  be  changed 
to  a  fractional  form,  and  the  division  performed  by  the  last  method. 
Hence  the  general 

Rule.  —  Invert  the  divisor,  and  then  proceed  as  in  multiplication  of 
fractions. 

Note  1.  —  When  either  divisor  or  dividend  is  a  whole  or  mixed  number,  or 
a  compound  fraction,  it  must  be  reduced  to  the  form  of  a  simple  fraction  be- 
fore dividing. 

Note  2.  —  Factors  common  to  both  numerator  and  denominator  should  be 
cancelled. 

Note  3.  —  When  the  given  fractions  have  a  common  denominator,  the  answer 
may  be  obtained  by  dividing  the  numerator  of  the  dividend  by  that  of  the 
divisor.  Also,  if  the  fractions  have  numerators  alike,  the  answer  may  be  ob- 
tained by  dividing  the  denominator  of  the  divisor  by  that  of  the  dividend. 

Note  4.  —  When  the  numerator  of  the  divisor  will  exactly  divide  the  numer- 
ator of  the  dividend,  and  the  denominator  of  the  divisor  exactly  divide  the 
denominator  of  the  dividend,  the  division  can  be  effected  in  that  way. 


COMMON  FRACTIONS.  175 

Examples. 

4.  Divide  T7<5-  by  &.  Ans.  6|. 

OPERATION. 

7  7  tf         62        6_2_ 

H)  "*"  62  =  l0  X  7"  =  T~0  =  6i  AnS- 
Or  A  ■*• *;- ^  -*-  10  -  6i  Ans. 

5.  Divide  2%  by  f .  Ans.  f . 

OPERATION. 

3 

28^7~^X0~4    AnS* 
4 

9         3         9-r-3  =  3A 
Or  —  -j —  =  —  —  Ans. 

28       7       28h-7=4 

6.  Divide  T7T  by  18.  Ans.  ^. 

7.  Divide  |  by  f  Ans.  f  f . 


8.  Divide  18  by  T7T. 

9.  Divide  ^T  by  f.  Ans.  £. 

10.  Divide  £$  by  f .  Ans.  f  —  If 

11.  Divide  ^  by  28.  Ans.  ^TV 

12.  Divide  TT  by  27. 

13.  Divide  T\  by  128.  Ans.  ^. 

14.  Divide  \\  by  98.  Ans.  T^. 

15.  Divide  £§■  by  19.  Ans.  £fc. 

16.  Divide  f  by  167.  Ans.  T^. 

17.  Divide  ££  by  49.  Ans.  TiJT. 

18.  Divide  y1^  by  15.  Ans.  ^is- 

19.  Divide  27  by  TT. 

20.  Divide  128  by  &.  Ans.  960. 

21.  Divide  98  by  Tf  Ans.  151TT. 

22.  Divide  19  by  Jf  Ans.  31Ty. 

23.  Divide  167  by  T|.  Ans.  200§. 

24.  Divide  49  by  £§. 

25.  Divide  15  by  TV  Ans.  225. 

26.  Divide  f  f  by  ^.  Ans.  4. 

27.  Divide  fy  by  ^  Ans.  3£f 

28.  Divide  £  by  f .  Ans.  -ff . 

29.  Divide  f|  by  T7T.  Ans.  lyW 


176  COMMON   FRACTIONS. 

30.  Divide  ^  by  |f 

31.  Divide  £f  by  TV  Ans.  llf 

32.  Divide  &  by  7J.  Ans.  T3TV 

33.  Divide  T8T  by  16f.  Ans.  ^V 

34.  Divide  llf  by  f 

35.  Divide  21f  by  18f  Ans.  l^J. 

36.  Divide  17T3T  by  28£f  Ans.  ff^-g. 

37.  Divide  161-&  by  14f  Ans.  llTJfT. 
88.  Divide  -fj  of  £  by  $  of  T8T.  Ans.  If 

39.  Divide  £  of  7T3T  by  T4T  of  17f  Ans.  f  f  J. 

40.  Divide  T<y  of  15  by  ^  of  22. 

41.  Bought  |-  of  a  coal-mine  for  $3675,  and  having  sold  -f 
of  my  share,  I  gave  §  of  the  remainder  to  a  charitable  society, 
and  divided  the  residue  among  7  poor  persons ;  what  was  the 
share  of  each  ?  Ans.  $  50  for  each  poor  person. 

42.  Of  an  estate  valued  at  $5000,  the  widow  receives  £, 
the  oldest  son  §  of  the  remainder;  the  residue  is  equally 
divided  among  7  daughters ;  what  is  the  share  of  each  daugh- 
ter? Ans.  $158|f. 

240.  When  the  dividend  is  a  mixed  number,  and  the  divi- 
sor a  whole  number,  we  may 

Divide  the  integral  part  of  the  mixed  number  as  in  division  of  whole 
numbers,  and  the  remainder  divide  as  in  Art.  239 ;  and  add  together 
the  results  for  the  quotient  required. 

Ex.  1.  Divide  27$  by  6.  Ans.  4f. 

OPERATION. 

6)27f 

4,  rem.  3f ;  Sft-^i  *£ x  6  =  if  =  f ;  4+f  =  4f,  Ans. 

2.  Divide  29f  by  9.  Ans.  3|f. 

3.  Divide  14£  by  7.  Ans.  2T^. 

4.  Divide  13f  by  8. 

5.  Divide  14f  by  6.  Ans.  2£g. 

6.  Divide  $  37f  among  9  men.  Ans.  $  4££. 

7.  Divide  $96§  among  11  persons.  Ans.  $8§|. 

8.  What  is  £  of  167T7T  cwt.  of  iron  ?  Ans.  20f  £  cwt 

9.  Divide  £  of  a  prize,  valued  at  $  1723,  equally  between 
12  seamen. 


COMMON   FRACTIONS.  177 

10.  What  will  a  barrel  of  flour  cost,  if  19  barrels  can  be  pur- 
chased for  $  107f  ?  Ans.  $  5.65 &. 

11.  If  15  pounds  of  raisins  can  be  obtained  for  $3f,  what 
will  1  pound  cost  ?  Ans.  $  0.21£f . 

12.  If  12  quarts  of  wine  cost  $3.75£,  what  will  a  quart 
cost? 

13.  If  $  19  will  buy  375|£  acres  of  land,  how  much  can  be 
bought  for  $  1  ?  Ans.  19§£f  acres. 

4- 

EEDUCTION  OF  COMPLEX  FRACTIONS. 

241.  A  complex  fraction  is  one  having  a  fraction  in  its 

3  21 

numerator  or  denominator,  or  in  both.     Thus,  -  and  -I  are 

i  f 

complex  fractions. 

242.  To  reduce  complex  to  simple  fractions. 

2 

Ex.  1.  Reduce  -  to  a  simple  fraction.  Ans.  Jf . 

"3" 

operation.  Since  the  numerator  of  a  fraction 

§•         „  1 6     A  *s  *^e  dividend,  and  the  denominator 

J  =  f  X  f  =  ff>  Ans.     t^e  diYisor  (Art.   216),  we  divide 
^  the  numerator,  f,  by  the  denomi- 

nator, -|,  as  in  division  of  fractions  (Art.  239). 

7 

2.  Eeduce  —  to  a  simple  fraction.         .       Ans.  ^  =  4J. 

If 

operation.  "We  reduce  the  nu- 

7      _  i         7         3  __  _   .       .           merator,    7,   and   the 

p  —  5  —  T  A    5  —  5-  —  *5>  ^nb-     denominator,    If,    to 

3  improper       fractions, 

and  then  proceed  as  in  Ex.  1 .     Hence,  to  reduce  complex  to  simple 

fractions, 

Consider  the  denominator  as  a  divisor,  and  the  numerator  as  a  divi- 
dend, and  proceed  as  in  division  of  fractions  (Art.  239). 

Note.  —  Another  and  often  a  ready  method  of  reducing  a  complex  fraction 
is  to  multiply  both  its  terms  by  the  least  common  multiple  of  their  denomi- 
nators. 

Examples, 
jl 

3.  Reduce  -^-  to  a  simple  fraction.  Ans.  ^. 

OPERATION. 

iXA-A=  ttV,  Ans. 


h    -y 


JL3   —  T   ^  TS  —  TZ  =  2S> 


178  COMMON   FRACTIONS, 


Or,  multiply  by  the  least  common  multiple  of  the  denominators, 

1  X  4  =2 

6£  X  4  =  26 


Ans. 


4f 
4.  Reduce  -£  to  a  simple  fraction.  Ans.  6T9¥. 


5.  Reduce  —-  to  a  simple  fraction. 


52 

7 

6.  Reduce  —  to  a  simple  fraction.  Ans.  1£ 

4| 

7.  Reduce   -—  to  a  simple  fraction.  Ans.  f  J. 

8 

6? 

8.  Reduce  -|  to  a  simple  fraction.  Ans.  § §. 

3 
2 

9.  Reduce   f  to  a  simple  fraction. 

"5 
Q 

10.  Reduce  -  to  a  whole  number.  Ans.  24. 

is 
i 

11.  Reduce   -  to  a  simple  fraction.  Ans.  f. 

2S 

51 

12.  Reduce  -^  to  a  mixed  number.  Ans.  12#. 

3 

13.  Reduce  ^  toa  simple  fraction. 

3 

14.  Reduce  — -   to  a  mixed  number.  Ans.  lh 

31 

15.  Reduce  ~  to  a  simple  fraction.  Ans.  -Jf. 

11  ? 

16.  Reduce   --— |  to  a  simple  fraction.  Ans.  f  f. 

17.  If  7  were  to  be  the  denominator  of  the  fraction  whose 

77 
numerator  is   -~>  what  would  be  its  value? 

U| 

18.  If  J  is  the  numerator  of  the  fraction  whose  denominator 
3. 

is  £  what  is  its  value  ?  Ans.  6£f . 

5 


COMMON   FRACTIONS.  179 

243.  Complex  fractions,  after  being  reduced  to  simple  ones, 
may  be  added,  subtracted,  multiplied,  and  divided,  according  to 
the  respective  rules"  for  simple  fractions. 

Examples. 

36  39 A 

1.  Add  i  of  f-  of  28^|  to  3—^.  Ans.  6-7V 

2.  Add  i,  2f,  —i  and  Q  together.     Ans.  3£t§$|$%|. 

495.  343 

3.  What  is  the  difference  between  — ~  and 


97  145T3T 

Ans.  Mfflfr. 

4.  What  is  the  continued  product  of  the  following  numbers 
27     87l     *    and  KfeV 


37|   98£    2i  128 

5.  Divide  §  of  7f  by  f  of  11T4T.  Ans.  |^. 

6.  Divide  f  of  91  by  T%  of  87.  Ans.  f  f  J?-. 


MISCELLANEOUS  EXAMPLES  IN  MULTIPLICATION  AND 
DIVISION  OF  FRACTIONS. 

1.  At  2f  bushels  to  an  acre,  how  many  bushels  of  wheat  will 
be  required  to  sow  7T4T  acres  ?  Ans.  17f . 

2.  Bought  8^-  bushels  of  apples  for  $  4.68f-;  what  did  they 
cost  per  bushel  ?  Ans.  $  0.57|. 

3.  Bought  a  bale  of  cloth  for  $  96f  ;  I  dispose  of  it  for  §  of 
the  cost,  and  by  so  doing  I  lose  $  2  on  a  yard ;  required  the 
number  of  yards  in  the  bale.  Ans.  18Tf^-yd. 

4.  If  a  dividend  be  18  J-  times  f  and  a  quotient  6 J-  times  §, 
what  was  the  divisor  ? 

5.  By  what  number  must  If  be  multiplied,  that  the  product 

o 

shall  just  equal  1  ?  Ans.  f . 

6.  Bought  a  horse  and  chaise  for  $  250,  and  paid  for  the 
harness  -fc  of  what  I  paid  for  the  horse.  The  chaise  cost  |4- 
the  value  of  the  horse.     What  was  the  price  of  each  ? 

Ans.  Horse,  $130£§;  chaise,  $119£f;  harness,  $83^1-. 


180  COMMON   FRACTIONS. 

7.  S.  Walker  has  engaged  to  work  at  yearly  wages  of  $  200 
and  a  suit  of  clothes.  At  the  end  of  9  months,  falling  sick,  and 
being  unable  to  labor  longer,  he  receives  the  suit  of  clothes  and 
$  144,  as  the  amount  justly  due.  What  was  the  cost  of  the 
clothes?  Ans.  $24. 

8.  What  will  be  the  result  if  £  of  f  of  3£  be  multiplied  by  £ 
of  itself,  and  the  product  divided  by  £  ? 

9.  Bought  13|  acres  of  land  at  $  25£  per  acre,  and  paid  for 
it  in  wheat  at  $  2-J  per  bushel.  How  many  bushels  did  it 
require  ?  Ans.  137££g  bushels. 

10.  How  long  will  it  take  a  man  to  travel  553  miles,  pro- 
vided he  travels  3 £  miles  per  hour,  and  9 £  hours  per  day  ? 

11.  If  $-2$  per  cord  is  paid  E.  Holmes  for  sawing  into 
three  pieces, wood  that  is  4  feet  long,  how  much  more  should 
he  receive  per  cord  for  sawing  into  pieces  of  the  same  length 
wood  that  is  8  feet  long  ?  Ans.  $  0.22£. 

12.  A  steamboat  leaves  New  Orleans,  January  1st,  bound  up 
the  river  to  a  place  distant  2317£  miles.  Her  forward  motion 
is  at  the  rate  of  9|  miles  per  hour  for  1GJ  hours  each  day, 
and  she  lies  at  anchor  in  the  night  for  fear  of  running  upon  a 
snag.  But  having  lost  her  anchor  on  the  fifth  day,  she  each 
succeeding  night  drifts  backward,  at  the  rate  of  2  miles  per 
hour.  On  what  day  of  January  will  she  reach  her  point  of 
destination  ?  Ans.  15th  day. 

A  PROPOSED  NUMERATOR,  OR  DENOMINATOR. 

244.  To  reduce  one  fraction  to  another  of  equal  value,  hav- 
ing a  proposed  numerator,  or  denominator. 

Ex.  1.  Eeduce  f  to  an  equivalent  fraction  having  4  for  a 
numerator.  Ans.  -^r-. 

operation.  The  proposed  numerator,  4,  is  such  a 

4  part  of  the  given  numerator  as  5  divided 

g"  X  $  by  4,  or  | .     Now,  as  the  numerator  pro- 

_ =  _  ^ns#  posed  is  only  |  as  large   as  the  given 

4  5f  numerator,  in  order  that  the  value  of  the 

5  X  '  two  fractions  be  the  same,  the  denomi- 

nator of  the  proposed  fraction  should  be 

only  £  as  large  as  the  denominator  of  the 

given  fraction.     Taking  -f  of  the  given  denominator,  7,  we  obtain 


COMMON    FRACTIONS.  >        181 

4 
5§,  which,  written  under  the  proposed  numerator,  gives  —  as  the 

fraction  required. 

2.  Eeduce  f  to  a  fraction  of  equal  value  having  12  for  a 

denominator. 

Since  the  proposed  denominator,  12, 

is  1£  of  the  given  denominator,  9,  we 

find  ^2-  of  the  given  numerator,  8,  for 

numerator  of  the  proposed  fraction ; 

J  2.  0f  8  =  101,  which,  written  over 

10# 
the  proposed  denominator,  gives  — $- 

l  u 

as  the  fraction  required. 

Rule.  —  Take  of  both  terms  of  the  given  fraction  such  a  fractional 
part  as  the  proposed  numerator,  or  denominator,  is  of  the  given  numer- 
ator, or  denominator,  and  the  result  will  be  the  required  fraction. 

Examples. 


OPERATION. 

12 

9 

X8 

103 

3 

Ans. 

12 

12 

T 

X0 

3.  Change  JJ  to  a  fraction  whose  numerator  shall  be  34. 

Ans.  f&. 

4.  Change  3f  to  a  fraction  whose  numerator  shall  be  9. 

Ans-i 

5.  Reduce  4  to  a  fraction  whose  numerator  shall  be  5. 

6.  Reduce  -ff  to  a  fraction  having  12  for  its  denominator. 

7.  Change  f  to  fifteenths.  Ans.  T9^. 

8.  Reduce  J  to  halves. 

33^ 

9.  Reduce  £$  to  thirty-fifths.  Ans.  ^~ 

DO 

\10.  J.  Holton  owns  -Jf-  of  a  wood-lot,  and  his  brother  ^/t  °f 
the  same  lot;  what  fraction  whose  denominator  shall  be  12 
will  express  the  part  each  owns  ?  Ans.  T4^. 

A  COMMON  NUMERATOR. 

245.  A  common  numerator  of  two  or  more  fractions  is 
a  common  multiple  of  their  numerators. 

246.  To  reduce  fractions  to  a  common  numerator. 

Ex.  1.  Change  f,  f,  f,  and  -^  to  other  fractions  of  the  same 
value,  having  a  common  numerator.         Ans.  f  |,  ff,  ff,  £#. 
N    16 


182  COMMON   FRACTIONS. 


OPERATION. 


36,  least  common  multiple  of  the  numerators,  =  common  nu- 
-3tj6-  of    4  =  48,  new  denominator,      f   =  f-§~]    [merator. 
^  of    5  mm  45^  new  denominator,      t   =  f  f 
ag.  of    7  =  42,  new  denominator,      f  —  ff  *  Ans# 
-8F6-  of  10  =  40,  new  denominator,     y9^  =  f^ 

"We  find  the  least  common  multiple  of  all  the  numerators,  which 
is  36,  for  the  common  numerator;  and  to  obtain  the  several  new  de- 
nominators we  take  such  a  part  of  the  given  denominators,  respec- 
tively, as  the.  common  numerator,  36,  is  of  each  given  numerator. 
Thus,  both  terms  of  each  fraction  being  proportionably  increased,  its 
value  is  not  changed. 

Rule.  —  Find  the  least  common  multiple  of  the  given  numerators  for 
a  common  numerator.  Take,  for  the  new  denominator  of  each  fraction, 
respectively,  such  a  part  of  its  given  denominator  as  the  common  numer- 
ator is  of  its  given  numerator. 

Note.  —  Compound  fractions,  or  whole  and  mixed  numbers,  must  be  re- 
duced to  simple  fractions,  and  all  to  their  lowest  terms,  before  finding  the 
common  numerator. 

Kxami-i.ks. 

2.  Reduce  f ,  % ,  £ ,  and  f  to  other  fractions  of  equal  value 
having  a  common  numerator.  Ans.  § $■,  f£,  §£,  §§. 

3.  Change  £,  2£,  and  1^  to  fractions  having  a  common  nu- 
merator. 

4.  A  can  travel  round  a  certain  island,  which  is  50  miles  in 
circumference,  in  4^  days,  B  in  G§  days,  and  C  in  G§  days. 
If  they  all  set  out  from  the  same  point,  and  travel  round  the 
island  the  same  way,  in  how  many  days  will  they  all  meet  at 
the  point  from  which  they  started,  and  how  many  times  will 
each  have  gone  round  the  island  ? 

Ans.  They  will  meet  in  320  days ;  A  will  have  gone  round 
the  island  75  times  ;  B,  50  times  ;  and  C,  48  times. 

GREATEST   COMMON  DIVISOR  OF  FRACTIONS. 

247  •  The  greatest  common  divisor  of  two  or  more  frac- 
tions is  the  greatest  number  that  will  divide  each  of  them,  and 
give  a  whole  number  for  the  quotient. 

248?  To  find  the  greatest  common  divisor  of  two  or  more 
fractions. 


COMMON   FUACTIONS.  183 

Ex.  1.  What  is  the  greatest  common  divisor  of  ^,  2§, 
and  5£.  Ans.  ■£%. 

OPERATION. 

4_    02     5i  JL_     20     16    =  J~2     100    _240 

Greatest  common  divisor  of  the  numerators  =    4  )  Greatest  com- 
Least  common  denominator  of  the  fractions  =  45  j     required. 

Having  reduced  the  fractions  to  equivalent  fractions  having  the 
least  common  denominator,  -we  find  the  greatest  common  divisor  of 
the  numerators  12,  100,  and  240  to  be  4.  Now,  since  the  12,  100, 
and  240  represent  forty-fifths,  their  greatest  common  divisor  is 
not  4,  a  whole  number,  bet  4  forty-fifths ;  therefore  we  write  the  4 
over  the  least  common  denominator,  45,  and  have  -£$  as  the  answer. 

Rule. — Reduce  the  fractions,  if  necessary,  to  their  least  common 
denominator.  The  greatest  common  divisor  of  the  numerators,  written 
over  the  least  common  denominator,  will  give  the  greatest  common  divisor 
required. 

Examples. 

2.  What  is  the  greatest  common  divisor  of  f ,  -f-,  f ,  and 
if?  An*.  #5, 

3.  What  is  the  greatest  common  divisor  of  12§,  9£,  and  8^? 

4.  What  is  the  greatest  common  divisor  of  ^-,  f ,  f- ,  and  f  ? 

Ans.  ^ 

5.  What  is  the  greatest  common  divisor  of  3f,  5T7^,  and  2^-? 

6.  A  farmer  has  33f-  bushels  of  corn,  67^-  bushels  of  rye, 
70  J  bushels  of  wheat.  He  wishes  to  put  this  grain,  without 
mixing,  into  the  largest  bags,  each  of  which  shall  contain  the 
same  quantity.  Required  the  number  of  bags  and  the  quantity 
each  will  contain. 

Ans.  The  capacity  of  each  bag,  3f  bushels ;  and  the  number 
of  bags,  51. 

7.  I  have  three  fields;  the  first  contains  73T7T  acres,  the 
second  88-^T  acres,  the  third  139|f  acres.  Required  the  lar- 
gest-sized house-lots  of  the  same  extent  into  which  the  three 
fields  can  be  divided,  and  also  the  number  of  lots. 

Ans.  Size  of  each  lot,  7T4T  acres  ;  number  of  lots,  41. 


LEAST   COMMON  MULTIPLE   OF  FRACTIONS. 

249.  The  least  common  multiple  of  two  or  more  fractions 
is  the  least  number  that  can  be  divided  by  each  of  them,  and 
give  a  whole  number  for  the  quotient. 


184  COMMON   FRACTIONS. 

250.  To  find  the  least  common  multiple  of  two  or  more 
fractions. 

Ex.  1.  What  is  the  least  common  multiple  of  f ,  T\,  and 
2T\?  Ans.  8  J. 

OPERATION. 

3       6       91     3     3     3.3 

Least  common  multiple  of  the  numerators  =  33  (  Least    com" 

.  .         „  .  —  =  84-  i  nion    multi- 

Greatest  common  divisor  of  denominators  =  4  4  V  pie  required. 

Having  reduced  the  fractions  to  their  simplest  form,  we  find  the 
least  common  multiple  of  the  numeratow,  3,  3,  and  33,  to  be  33. 
Now,  since  the  3,  3,  and  33  are,  from  the  nature  of  a  fraction,  divi- 
dends, of  which  their  respective  denominators,  4,  8,  and  16,  are  the 
divisors  (Art.  216),  the  least  common  multiple  of  the  fractions  is  not 
33,  a  whole  number,  but  so  many  fractional  parts  of  the  greatest 
common  divisor  of  the  denominators.  This  common  divisor  we  find 
to  be  4,  which,  written  as  the  denominator  of  the  33,  gives  ^  =  8 \ 
as  the  least  number  that  can  be  exactly  divided  by  the  given  fractions. 

Rule.  —  Reduce  the  fractions,  if  necessary,  to  their  lowest  terms. 
Then  find  the  least  common  multiple  of  the  numerators,  which,  written 
over  the  greatest  common  divisor  of  the  denominators,  will  give  the  least 
common  multiple  required. 

Note.  —  The  least  whole  number  that  will  contain  two  or  more  fractions 
an  exact  whole  number  of  times,  is  the  least  common  multiple  of  their  numera- 
tors. Thus,  4,  the  least  common  multiple  of  the  numerators  of  |  and  |,  is  the 
least  whole  that  can  be  divided  by  those  fractions,  and  give  a  whole  number 
for  a  quotient;  4  -r  §  =  6 ;  4-7-5  =  5. 

Examples. 

2.  What  is  the  least  common  multiple  of  f ,  f ,  and  f  ? 

Ans.  2±  =  24. 

3.  Find  the  least  number  that  3^f ,  7£,  and  5J  will  divide 
without  a  remainder.  Ans.  15  J . 

4.  What  is  the  least  common  multiple  of  §,  f ,  and  T9^  ? 

5.  What  is  the  smallest  sum  of  money  with  which  I  could 
purchase  a  number  of  sheep  at  §  2£  each,  a  number  of  calves 
at  $  4£  each,  and  a  number  of  yearlings  at  $  9f  each  ?  and 
how  many  of  each  could  I  purchase  with  this  money  ? 

Ans.  $  112 \  ;  50  sheep  ;  25  calves  ;  12  yearlings. 

6.  There  is  a  certain  island  80  miles  in  circumference.  A, 
B,  and  C  agree  to  travel  round  it.  A  can  walk  3 J  miles 
in  an  hour,  B  4f  miles,  and  C  h\  miles.  They  start  from 
the  same  point  and  travel  round  the  same  way,  and  continue 


COMMON    FRACTIONS.  185 

their  travelling  8  hours  a  day,  until  they  shall  all  meet  at  the 
point  from  which  they  started.  In  how  many  days  will  they 
all  meet,  and  how  far  will  each  have  travelled  ? 

Ans.   In  17-f  days  ;  A  480m.,  B  640m.,  and  C  720m. 
7.  How  many  times  the  least  common  multiple  of  3J,  4|, 
and  5£,  is  the  least  whole  number  that  3^,  4§,  and  5£  will 
exactly  divide. 

DENOMINATE  FRACTION. 

251.  A  Denominate  Fraction  is  one  in  which  the  unit 
of  the  fraction  is  a  denomination  of  a  compound  number ;  as, 
§  of  a  pound,  |-  of  a  mile,  and  I  of  a  gallon. 

REDUCTION  OF  DENOMINATE  FRACTIONS. 

252.  Reduction  of  denominate  fractions  is  the  process 
of  changing  fractions  from  the  unit  of  one  denomination  to 
that  of  another,  without  altering  their  value. 

253.  To  reduce  a  denominate  fraction  from  a  higher  de- 
nomination to  a  lower. 

Ex.  1.     Reduce  ^J^  of  a  pound  to  a  fraction  of  a  penny. 

Ans.  f  d. 

OPERATION. 

1  X  20  =  20    20  X  12  =  240  ,  _  3  ,  A 
640       640s' 5  640       640  ~  8 

3 

n„      1     X^X^  =  3,    .a        Since  20s.  make  apound, 

0?0  8  ttere  wil1  be  20  times  as 

-^       ^  many  shillings  as  pounds, 

*^  or  -g24°o-s. ;  and  since  12d. 

make  a  shilling,  there  will  he  12  time's  as  many  pence  as  shillings, 

orff£d.  =  fd. 

Rule.  —  Multiply  the  given  fraction  by  the  same  numbers  that  would 
be  employed  in  the  reduction  of  whole  numbers  to  the  lower  denomina- 
tion required. 

Examples. 

2.  Reduce  y^W  of  a  pound  to  the  fraction  of  a  farthing. 

3.  Reduce  ^(j  of  a  pound  troy  to  the  fraction  of  a  grain. 

4.  Reduce  2  As"  °f  a  pound,  apothecaries'  weight,  to  the 
fraction  of  a  scruple. 

5.  Reduce  -s$j$  of  a  cwt.  to  the  fraction  of  an  ounce. 

16* 


186  COMMON   FRACTIONS. 

6.  Reduce  ^y^u"  of  a  ton  to  the  fraction  of  a  pound. 

7.  What  part  of  an  inch  is  ■%£ ^  of  an  ell  English  ? 

8.  What  part  of  an  inch  is  TTuVsTy  of  a  mile  ? 

9.  Reduce  ^(jtf  oi"  a  league  to  the  fraction  of  an  inch. 

10.  Reduce  3-5  <j^3<y-5su-  of  an  acre  to  the  fraction  of  an  inch. 

11.  Reduce  yyV?  of  a  tun  of  wine  measure  to  the  fraction 
of  a  quart. 

12.  What  part  of  a  pint  is  -^f  ^  of  a  bushel  ? 

13.  What  part  of  a  minute  is  ^^Hbit  0I>  a  year  ? 

14.  Reduce  ^^tt  °f  a  hundred-weight  to  a  fraction  of  an 
ounce. 

254.     To  reduce  a  denominate  fraction  from  a  lower  de- 
nomination to  a  higher. 

Ex.  1.     Reduce  §  of  a  penny  to  a  fraction  of  a  pound. 

Ans.  s£<j. 

OPERATION. 

3  00  -C  ■€       A 

8X12  =  96  S'5  96  X  20  =  im    '  "  640  S' 

(^     #  1      r      a  Since  12  pence  make 

'  8X^X20  =  640  *  shillinS>  tkere  yill  be 

7  TJ  as  man)r  shillings  as 

4  pence,    or    -As.  ;     and 

since  20s.  make  a  pound,  there  will  be  -fa  as  many  pounds  as  shil- 
lings, or  ^  £.    Ans. 

Rule.  —  Divide  the  fraction  by  the  same  numbers  that  would  be  em- 
ployed in  the  reduction  of  whole  numbers  to  a  higher  denomination. 

Examples. 

2.  What  part  of  a  pound  is  f  of  a  farthing  ? 

3.  What  part  of  a  pound  is  f  of  a  grain  troy  ? 

4.  Reduce  |-  of  a  scruple  to  the  fraction  of  a  pound. 

5.  Reduce  T6T  of  an  ounce  to  the  fraction  of  a  hundred- 
weight. 

6.  Reduce  £  of  a  pound  to  the  fraction  of  a  ton. 

7.  Reduce  £  of  an  inch  to  the  fraction  of  an  ell  English, 

8.  Reduce  f  of  an  inch  to  the  fraction  of  a  mile. 

9.  Reduce  ^  of  an  inch  to  the  fraction  of  a  league. 
10.  Reduce  f  of  an  inch  to  the  fraction  of  an  acre. 


COMMON   FRACTIONS.  187 

11.  Reduce  J  of  a  quart  to  the  fraction  of  a  tun,  wine  measure. 

12.  Reduce  f  of  a  pint  to  the  fraction  of  a  bushel. 

13.  Reduce  ^  of  a  minute  to  the  fraction  of  a  year  (365£ 
days). 

14.  "What  part  of  a  hundred-weight  is  f  of  an  ounce  ? 

Ans.  sgjjxi' 
255 .     To  find  the  value  of  a  fraction  in  whole  numbers  of 
lower  denominations. 

Ex.  1.  What  is  the  value  of  T3T  of  a  £.     Ans.  5  s.  5d.  lT9Tfar. 

OPERATION. 

3£. 
20 


1  1 )  6  0  s.  (5  s.  Since  l£.  =  20s.,  A-  of  a  £.  is  ■& 

5  5  of  20s.  =  ff  s.  =  5Xs. ;  and  since 

~~ 5S#  Is.  =  12d.,  T5T  of  a  shilling  is  T5T  of 

I  o   *  12d.  =  ffd.  =  5T5Td. ;  and,  since  Id. 

— =  4far . ,  T5T  of  a  penny  =  T5T  of  4far. 

1  1  )  6  0  d.  (  5  d.  =  |f far.  =  lTVar.    Therefore,  T\£. 

5  5  =  5s.  5d.  lT9Tfar.     This  is  equiva- 
lent to  multiplying  the  numerator  of 


5d. 


the  fraction  by  the  numbers  required 


4  to  reduce  it  to  successive  lower  de 


11)20  far  (  l-9  far  nominations,  beginning  with  the  high- 

i  i    '  ll       '  est,  and  dividing  each  product  by 

the  denominator,  as  in  the  operation. 

T9Tfar. 

Ans.  5s.  5d.  lT9Tfar. 

Rule.  — Multiply  the  numerator  of  the  given  fraction  by  the  number 
required  to  reduce  it  to  the  next  lower  denomination,  and  divide  the 
product  by  the  denominator. 

TJien,  if  there  is  a  remainder,  proceed  as  before,  until  it  is  reduced 
to  the  denomination  required. 

Examples. 

2.  What  is  the  value  of  fa  of  a  shilling  ?  Ans.  3£d. 

3.  What  is  the  value  of  J  of  a  guinea,  at  28  shillings  ? 

Ans.  21s.  9d.  l^far. 

4.  What  is  the  value  of  T7T  of  a  cwt.  ? 

5.  What  is  the  value  of  f  of  a  lb.  avoirdupois  ? 

Ans.  7oz.  ljdr. 

6.  What  is  the  value  of  §  of  a  lb.  troy  ? 

Ans.  lOoz.  13pwt.  8gr. 


188  COMMON   FRACTIONS. 

7.  What  is  the  value  of  -fa  of  a  lb.  apothecaries'  weight  ? 

Ans.  35   55  19  12T4jgr. 

8.  What  is  the  value  of  f  of  an  ell  English  ? 

Ans.  2qr.  3na.  Ojin. 

9.  What  is  the  value  of  -f£  of  a  mile  ? 

Ans.  6fur.  30rd.  12ft.  8T43-in. 

10.  What  is  the  value  of  f  of  a  furlong  ? 

Ans.  35rd.  9ft.  2in. 

11.  What  is  the  value  of  fa  of  an  acre  ? 

Ans.  2R.  6rd.  4yd.  5ft.  127Tyn. 

12.  What  is  the  value  of  T9T  of  a  rod  ? 

13.  What  is  the  value  of  TVj  of  a  cord  ? 

Ans.  9ft.  1462TVn. 

14.  What  is  the  value  of  T^-  of  a  hhd.  of  wine  ? 

Ans.  6gal.  2qt.  lpt.  0T%gi. 

15.  What  is  the  value  of  £  of  a  hhd.  of  beer  ?     Ans.  42gal. 

16.  What  is  the  value  of  ££  of  a  year  (365  J  days)  ? 

Ans.  174d.  16h.  26m.  5^sec. 

3  3 

17.  What  is  the  value  of  7-£  of  a  dollar? 

256#  To  find  the  value  of  whole  numbers  in  a  fraction  of  a 
higher  denomination. 

Ex.  1.  What  part  of  a  £.  are  5s.  5d.  1^-far.  ? 

operation.  We   reduce   the   3s. 

5s.  5d.  lT9Tfar.  =    2880  3  5d.  l^tkr.  to  elevenths 

2£  __  10560  —  TT  of  a  farthing,  the  low- 

est denomination  in  the 
question,  for  the  numerator  of  the  required  fraction,  and  1  £.  to  the 
same  denomination  for  the  denominator.  We  then  have  the  fraction 
Toi^o^"  ==*-$£&  as  the  answer. 

Rule.  — Reduce  the  given  numbers  to  the  lowest  denomination  men- 
tioned in  either  of  them.  Then  write  the  number  which  is  the  fractional 
part  for  the  numerator,  and  the  other  number  for  the  denominator,  of 
the  required  fraction. 

Examples. 

2.  Reduce  3^-d.  to  the  fraction  of  a  shilling.         Ans.  ^fc. 

3.  Reduce  21s.  9d.  l^far.  to  the  fraction  of  a  guinea. 

Ans.  J. 


COMMON   FRACTIONS.  189 

4.  Reduce  2qr.  131b.  lOoz.  2££dr.  to  the  fraction  of  a  cwt. 

Ans.  T7T. 

5.  What  part  of  a  pound  avoirdupois  are  7oz.  ljdr.  ? 

6.  What  part  of  a  pound  troy  are  lOoz.  13pwt.  8gr.  ? 

Ans.  f . 

7.  Express  1  gallon  liquid  measure  as  a  fraction  of  a  gallon 
dry  measure.  Ans.  ^f  f  f . 

8.  What  part  of  a  yard  are  2qr.  Ona.  lT53in.  ?        Ans.  T7^. 

9.  What  part  of  an  ell  English  are  2qr.  3na.  0£in.  ? 

Ans.  f. 

10.  What  part  of  a  mile  are  6fur.  30rd.  12ft.  8^11.  ? 

11.  Reduce  35rd.  9ft.  2in.  to  the  fraction  of  a  furlong. 

Ans.  f. 

12.  What  part  of  an  acre  are  2R.  6rd.  4yd.  oft.  127-^. 

Ans.  TV 

13.  What  part  of  a  square  rod  are  144ft.  19T17in.  ? 

Ans.  -9-. 

14.  What  part  of  a  cord  are  9ft.  1462T2^in.  ? 

15.  What  part  of  a  hogshead  of  wine  are  6gal.  2qt.  lpt. 
0T%gi.?  Ans.TV 

16.  What  part  of  a  pound  avoirdupois  is  1  pound  troy  ? 

Ans.  iff. 

17.  What  part  of  a  year  (365J  days)  are  174d.  16h.  26m. 
5^53  sec.  ?  Ans.  J£. 


ADDITION  OF  DENOMINATE  FRACTIONS. 

257.     To  add  denominate  fractions. 
Ex.  1.  Add  T7^  of  a  £.  to  T9T  of  a  £. 

Ans.  1£.  7s.  Id.  2-^far. 

FIRST    OPERATION.  ,     ^  ^  ^  ^  °{^ 

X  of  a  £.  =  10s.  9d.  O^ffar.  ^act\on  separately  and  add 

the    two   values    together, 


T9T  of  a  £.  =  16s.  4d.  1-^far.  according   to  the   rule  for 

Ans.  1£.  7s.  Id.  2T^far.       ^^^I?^111"1  numbers* 


SECOND    OPERATION. 


A£-  +  tt£-  =  tB£-  —  1£-  7s.  Id.  2T%far.  Ans. 

By  the  second  operation,  we  first  add  the  two  fractions  together, 
and  then  find  the  value  of  their  sum.     (Art.  255.) 


190  COMMON   FRACTIONS. 

Examples. 

2.  Add  together  -^T  of  a  ton  and  £  J  of  a  cwt. 

Ans.  13cwt.  2qr. 

3.  Add  together  f  of  a  yard,  \  of  an  ell  English,  and  f-  of 
a  qr. 

4.  Add  together  T\-  of  a  mile,  -£$  of  a  furlong,  and  ^  of  a 
yard.  Ans.  5fur.  16rd.  Oft.  3-j^in. 

5.  A  has  three  house-lots ;  the  first  contains  £  of  an  acre, 
the  second  §  of  an  acre,  and  the  third  -£J  of  an  acre.  How 
many  acres  do  they  all  contain  ? 

Ans.  2A.  1R.  9p.  142ft.  87fin. 

6.  A  man  travelled  18^  miles  the  first  day,  23-j-^  miles  the 
second  day,  and  192T  miles  the  third  day.  How  far  did  he 
travel  in  the  three  days  ?         Ans.  61m.  2fur.  3rd.  13ft.  4f  in. 

7.  Add  \±  of  a  gallon  of  wine  to  y1^  of  a  hhd. 

8.  Add  y\  of  a  week  to  -J  of  a  day.  Ans.  2d.  9h.  18m. 

9.  Add  £  of  a  square  foot  to  J  a  foot  square.       Ans.  1  foot. 
10.  Add  G  inches  to  llrd.  16ft.  oin.         Ans.  12rd.  Oft.  oin. 

SUBTRACTION  OF  DENOMINATE  FRACTIONS. 

258 •     To  subtract  one  denominate  fraction  from  another. 
Ex.  1.     From  -fa  of  a  £.  take  f  of  a  £. 

Ans.  8s.  3d.  ljffar. 

fiust  operation.  We  find  the  value  of  each  fraction 

-^Y  of  a  £.  =  12s.  8T8Td.  separately,  and  subtract  one  from  the 

2    of  a  £.  =     4s.  5  }  d.  other,  according  to  the  rule  for  sub- 

— — —  tracting  compound  numbers.     (Art. 

Ans.    8s.  3£§d.  146.) 

SECOND    OPERATION.  By     tllC     SCCOnd 

T7T£.  —  f £.  =  f ^£.  =  8s.  3d.  l-J-ffar.  Ans.     operation,  we  first 

subtract  the  less 
fraction  from  the  greater,  and  then  find  the  value  of  their  difference. 
(Art.  255.) 

Examples. 

2.  From  %  of  an  ell  English  take  f  of  a  yard. 

Ans.  3qr.  Ona.  2-^in. 

3.  Take  -fy  of  a  furlong  from  §  of  a  mile. 

Ans.  lfur.  5rd.  10ft.  lOin. 


COMMON   FRACTIONS.  191 

4.  Take  -f  of  a  mile  from  f  of  a  degree. 

Ans.  48m.  6fur.  17rd.  8ft.  7f-in. 

5.  From  T4T  of  an  acre  take  f  of  a  rod. 

Ans.  1R.  17p.  22yd.  2ft.  108in. 

6.  From  T9^  of  a  cord  take  -fT  of  a  cord. 

7.  From  t7tj  of  a  hogshead  of  wine  there  leaked  out  $  of  it ; 
what  remained  ?  Ans.  6gal.  3qt.  Opt.  1-g^-gi. 

8.  From  Boston  to  Concord,  N.  H.,  the  distance  is  72  miles ; 
f  of  this  distance  having  been  travelled,  how  much  remains  ? 

Ans.  30m.  6fur.  34rd.  4ft.  8f-in. 

9.  From  f  of  a  year  take  f  of  a  week. 

Ans.  lOlda.  5h.  54m.  17|sec. 

10.  From  T4r  of  an  acre  take  J  of  a  foot. 

MISCELLANEOUS  EXAMPLES  IN  FRACTIONS. 

1.  How  far  will  a  man  walk  in  17T3T  hours,  provided  he 
goes  at  the  rate  of  4 J-  miles  an  hour  ? 

Ans.  82m.  4fur.  8rd.  1ft.  4in. 

2.  How  much  land  is  there  in  a  field  which  is  29t7tT  rods 
square  ?  Ans.  5A.  1R.  32p.  141ft.  109f|fin. 

3.  How  much  wood  in  a  pile  which  is  17J  feet  long,  7-jL- 
feet  high,  and  4f  feet  wide  ?  Ans.  4C.  66|$Jft. 

4.  What  is  the  value  of  19  J  barrels  of  flour,  at  $  6f  a  bar- 
rel ? 

o.  What  is  the  value  of  376|£  acres  of  land,  at  $75§  per 
acre?.  Ans.  $28387.06£. 

6.  What  cost   17-j2j3j  quintals  of  fish,  at  $  4.75  per  quin- 
tal ?  Ans.  $  81.55T6T52-. 

7.  What  cost  1670T%  pounds  of  coffee,  at  12f  cents  per 
pound?  Ans.  $212.99|f. 

8.  What  cost  28^T  tons  of  Lackawana   coal,  at   $llf   a 
ton  ?  Ans.  $  333.27T3T. 

9.  Bought  37^  hogsheads  of  molasses,  at  $  17.62J  a  hogs- 
head ;  wThat  was  the  whole  cost  ?  Ans.  $  655.202-5T. 

10.  What  cost  £  of  a  cord  of  wood,  at  $  5.75  a  cord  ? 

11.  What  are  the  contents  of  a  field  which  is  139f-  rods 
long,  and  38$  rods  wide  ?  Ans.  33A.  3R.  15£f p. 

12.  Bought  15  loads  of  wood,  each  containing  llf  feet,  cord 


192  COMMON   FRACTIONS. 

measure.     I  divide  it  equally  between  9  persons ;  what  does 
each  receive  ?  Ans.  19-Lft. 

13.  If  the  transportation  of  18f  tons  of  iron  costs  $48.15f, 
what  is  it  per  ton  ?  Ans.  $  2.62^%. 

14.  If  a  hogshead  of  wine  costs  $  98  J,  what  is  the  price  of 
one  gallon  ? 

15.  If  5  bushels  of  wheat  cost  $8§,  what  will  a  bushel  be 
worth  ?  Ans.  $  1.64|. 

16.  What  will  11  hogsheads  and  17J  gallons  of  wine  cost, 
at  19f  cents  a  gallon  ?  Ans.  $  140.32J . 

17.  How  many  bottles,  each  containing  If  pints,  are  suffi- 
cient for  bottling  a  hogshead  of  cider  ?  Ans.  288. 

18.  I  have  a  shed  which  is  18T7j  feet  long,  10^  feet 
wide,  and  7^4  feet  high ;  how  many  cords  of  wood  will  it  con- 
tain ?  Ans.  1 1 C.  1 24T%9Fft. 

19.  What  will  G|  pounds  of  tea  cost,  at  65  J  cents  per 
pound?  Ans.  $4.52^2- 

20.  How  many  cubic  feet  does  a  box  contain,  that  is  8f  feet 
long,  5T72-  feet  wide,  and  8  feet  high  ?  Ans.  146T9^ft. 

21.  How  many  feet  of  boards  will  it  take  to  cover  a  side  of 
a  house  which  is  46^-  feet  long  and  17£  feet  high  ? 

22.  Required  the  number  of  square  feet  on  the  surface  of  7 
boxes,  each  of  which  is  5£  feet  long,  2T^-  feet  high,  and  3T\j 
feet  wide ;  required  also  the  number  of  cubic  feet  they  would 
occupy  ?  Ans.  527f  f  ft. ;  286f  f  $  cubic  feet. 

23.  A  certain  room  is  12  feet  long,  11£  feet  wide,  and  7 J 
feet  high ;  how  much  will  it  cost  to  plaster  it,  at  2f  cents  per 
square  foot  ?  Ans.  $  13.48  J. 

24.  A  man  has  a  garden  that  is  14 J  rods  long,  and  10£  rods 
wide ;  he  wishes  to  have  a  ditch  dug  around  it,  that  shall  be 
3  feet  wide  and  4£  feet  deep ;  what  will  be  the  expense,  if  he 
give  2  cents  per  cubic  foot  ?  Ans.  $  223.7 6£. 

25.  How  many  bushels  of  grain  will  a  box  contain  which  is 
14T72-  feet  long,  5-}-^-  feet  deep,  and  4 J  feet  wide,  there  being 
2150!  cubic  inches  in  a  bushel  ?  Ans.  294|f  f  f  bu. 

26.  Which  will  contain  the  most,  and  by  how  much,  a  box 
that  is  10  feet  long,  8  feet  wide,  and  6  feet  deep,  or  a  cubical 
one,  each  of  whose  sides  measures  8  feet  ? 

Ans.  The  last  contains  32  cubic  feet  the  most. 


COMMON   FRACTIONS.  193 

27.  Which  will  contain  the  most  gallons,  a  cistern  that  is  7£ 
feet  long,  6  feet  wide,  and  5J  feet  deep,  or  one  that  is  9^-  feet 
long,  4J-  feet  wide,  and  5J-  feet  deep  ? 

Ans.  The  first  cistern  contains  92|-  gallons  most. 

28.  My  field  has  four  sides.  The  first  side  is  31  rods  13^ 
feet  in  length ;  the  second,  41  rods  ly9^  feet ;  the  third,  38 
rods  0-i  feet;  and  the  fourth,  45  rods  12T7^-  feet.  I  wish  to 
enclose  this  field  with  a  rail-fence  four  rails  high,  using  rails  of 
equal  length.  Required  the  length  of  the  longest  rails  that  can 
be  used,  allowing  that  the  rails  lap  by  each  other  y7^  of  a  foot ; 
also  the  number  of  rails  it  will  take  to  fence  it. 

Ans.  Length,  13-|-  feet ;  number,  808. 

29.  Eequired  the  least  number  of  yards  of  velvet,  expressed 
by  a  whole  number,  that  can  be  cut  up  without  waste,  into  vest 
patterns  of  f ,  f ,  or  T  yards  each  ?  Ans.  30  yards. 

30.  If  a  company  whose  capital  stock  is  divided  into  100 
equal  shares  should  conclude  to  divide  the  same  stock  into 
only  30  shares,  how  much  larger  would  a  share  of  the  latter 
be  than  one  of  the  former  size  ? 

31.  D.  Ripley's  farm  contains  31  A.  3R.  6 p.,  and  J.  Ford's 
farm  contains  39A.  2R.  37^-p.  What  is  the  fraction,  in  its  sim- 
plest form,  that  will  express  their  comparative  size  ?  Ans.  f . 

32.  Bought  68  barrels  of  flour,  at  $  1\\  per  barrel ;  what 
was  the  amount  of  the  whole  ?  Ans.  $  538  J. 

33.  What  cost  8§  acres  of  land,  at  $  42f  per  acre  ? 

Ans.  $369.20. 

34.  How  shall  four  3's  be  arranged,  that  their  value  shall  be 
nothing  ? 

35.  I  have  a  room  20  feet  long,  15  feet  wide,  and  8J  feet 
high.  This  room  contains  4  windows,  each  of  which  is  5£ 
feet  in  height  and  3-J-  feet  in  width.  There  are  two  doors  7 
feet  high  and  3  feet  wide.  The  mop-boards  are  §  of  a  foot 
wide.  A  mason  has  agreed  to  plaster  this  room  at  6|-  cents 
per  square  yard ;  a  painter  is  to  lay  on  the  paper  at  9  cents 
per  square  yard ;  the  paper  which  I  wish  to  have  laid  on  is  2f 
feet  wide,  for  which  I  pay  5  cents  per  yard.  What  is  the 
amount  of  my  bill  for  plastering,  for  papering,  and  for  paper  ? 

Ans.    For  plastering,   $5.1  Iff;  for  papering,   $4.37;  for 
paper,  $  2.80^ 
N     17 


194  COMMON   FRACTIONS, 


EXAMPLES  TO  BE  PERFORMED  BY  ANALYSIS. 

1.  If  -J-  of  a  bushel  of  corn  cost  63  cents,  what  cost  a 
bushel  ?  what  cost  15  bushels  ?  Ans.  $  10.80. 

Illustration.  —  If  7  eighths  of  a  bushel  cost  63  cents,  1 
eighth  will  cost  1  seventh  of  63  cents  =  9  cents ;  and  8  eighths 
will  cost  8  times  9  cents  —  72  cents,  and  15  bushels  will  cost 
15  times  72  cents  —  $  10.80. 

2.  If  4 Jib.  of  pepper  cost  $  2.15,  what  cost  1  pound  ?  what 
cost  301b.  ?  Ans.  $  13.50. 

Illustration.  —  In  4£lb.  there  are  ^lb.  Then,  if  43 
ninths  lb.  cost  $2.15,  1  ninth  will  cost  1  forty-third  of  $2.15 
=  $  0.05,  and  9  ninths  or  lib.  will  cost  9  times  $  0.05  —  $  0.45, 
and  301b.  will  cost  30  times  $0.45  =  $  13.50. 

3.  When  $  1728  are  paid  for  30T\  tons  of  iron,  what  cost  1 
ton  ?  what  cost  7££  tons  ?  Ans.  $  432. 

4.  When  $  432  are  paid  for  7^J  tons  of  iron,  what  quantity 
should  be  received  for  $  1728  ? 

5.  For  7 -J- J  tons  of  iron  there  were  paid  $  432  ;  what  sum 
will  it  require  to  pay  for  30^  tons  ? 

6.  For  30T^  tons  of  iron  S  1728  were  paid;  what  quantity 
should  be  received  for  $  432  ? 

7.  Gave  7^  bushels  of  rye  for  a  barrel  of  flour  ;  how  much 
rye  will  it  then  require  to  purchase  6|  barrels  of  flour? 

Ans.  49  J|-  bushels. 

8.  Divide  $  1728  among  17  boys  and  15  girls,  and  give  each 
boy  -^j  as  much  as  a  girl ;  what  sum  will  each  receive  ? 

Ans.  Each  girl,  $  66f  f ;  each  boy  $  42f  f . 

9.  If  £  of  a  ton  of  hay  cost  $  14.49,  what  cost  4f  tons  ? 

10.  If  44  tons  of  hay  cost  $  82.50^,  what  part  of  a  ton  will 
$  14.49  buy  ? 

11.  If  $  14.49  will  buy  J  of  a  ton  of  hay,  how  much  hay  can 
be  obtained  for  $  82.50f  ? 

12.  When  $  82.50"f  are  paid  for  4f  tons  of  hay,  what  will  be 
the  cost  of  J  of  a  ton  ? 

13.  When  14 J  tons  of  copperas  are  sold  for  $500,  what  is 
the  value  of  1  ton  ?  what  is  the  value  of  9f£  tons  ? 


COMMON   FRACTIONS.  195 

14.  When  9-J£  tons  of  copperas  are  sold  for  $  333.33£,  what 
is  the  value  of  14|-  tons  ? 

15.  Gave  $  333.33^-  for  9f  i  tons  of  copperas  ;  what  quantity 
of  copperas  should  be  received  for  $  500  ? 

'16.  For  14§-  tons  of  copperas  $500  were  paid  ;  how  much 
might  be  purchased  for  $  333.33^  ? 

17.  Purchased  97J  gallons  of  molasses  for  $31.32;  what 
cost  1  gallon  ?  what  cost  763-g-  gallons  ? 

18.  Sold  763f  gallons  of  molasses  for  $  244.36  ;  what  should 
I  receive  for  97-J  gallons  ? 

19.  If  $244.36  will  buy  763f  gallons  of  molasses,  what 
quantity  can  be  obtained  for  $  31.32  ? 

20.  Gave  19751b.  of  flax  for  40  barrels  of  flour;  how  many 
pounds  were  given  for  1  barrel  ?  how  many  pounds  would  it 
require  to  buy  144  barrels  ?  Ans.  71101b. 

21.  If  17  bushels  of  rye  cost  $  15.75,  what  cost  1  bushel  ? 
what  cost  9 1  bushels  ?  Ans.  $  8.5  6f  J. 

22.  If  9  barrels  of  flour  cost  $  5 Of- ,  what  cost  1  barrel  ?  what 
cost  87f  barrels  ?  Ans.  $  492^.   . 

23.  If  13  boarders  consume  a  barrel  of  pork  in  78  days,  how 
long  would  it  last,  if  7  more  boarders  were  added  to  their  num- 
ber ?  Ans.  50^  days. 

24.  If  a  man  by  laboring  10  hours  a  day  can,  in  9  days,  per- 
form a  certain  piece  of  work,  how  many  days  would  it  require 
to  do  the  same  work,  were  he  to  labor  15  hours  a  day  ? 

25.  If  a  man,  by  laboring  15  hours  a  day,  in  6  days  can  per- 
form a  certain  piece  of  work,  how  many  days  would  it  require 
to  do  the  same  work  by  laboring  10  hours  a  day  ? 

26.  If  a  man,  by  laboring  10  hours  a  day,  can  in  9  days  per- 
form a  certain  piece  of  work,  how  many  hours  must  he  labor 
each  day  to  perform  the  same  work  in  6  days  ? 

27.  Sold  17T3T  bushels  of  corn  for  $  5f-  ;  what  was  received 
for  1  bushel  ?  what  should  I  have  charged  for  97f  bushels  ? 

Ans.  $30fff. 

28.  Bought  9f  tons  of  hay  at  $  19$  per  ton;  for  what  must 
it  be  sold  per  cwt.  to  gain  $  7  on  my  bargain  ?  Ans.  $  l^ff  ^. 

29.  If  I  sell  hay  at  $  If  per  cwt.,  what  should  I  give  for  9f 
tons,  that  I  may  make  $  7  on  my  bargain  ?  Ans.  $  329. 

30.  How  many  bushels  of  corn  at  $0.75  per  bushel  will 


196  COMMON   FRACTIONS. 

it  require  to  purchase  47T3T  bushels  of  wheat   at   $2§  per 
bushel?  Ans.  168g%bu. 

31.  If  15  cords  of  wood  cost  $  57T9T,  what  cost  1  cord?  what 
cost  19  J  cords  ? 

32.  If  19 1  cords  of  wood  cost  $  76T6ft,  how  many  cords  may- 
be obtained  for  $  57T9T  ? 

33.  At  7T3IJ  shillings  per  yard,  what  cost  47-J  yards  ? 

Ans.  17£.  5s.  6§d. 

34.  When  172£.  15s.  Ofd.  are  paid  for  47£  yards  of  broad- 
cloth, what  is  the  value  of  1  yard  ?       Ans.  3£.  12s.  llfffd. 

35.  If  lib.  of  sugar  cost  ^  of  a  dollar,  what  is  the  value  of 
43flb.?  Ans.  $  23.61ft. 

36.  If  17glb.  of  sugar  cost  $  2T7T,  what  cost  501b. 

Ans.  $  7.58-HhB • 

37.  Bought  87^  yards  of  broadcloth  for  $  612  ;  what  was  the 
value  of  14ft-  yards  ?  Ans.  $  102.90. 

38.  If  £  of  an  acre  of  land  cost  $  43.75,  what  cost  10  acres  ? 

39.  When  $  500  are  paid  for  10  acres  of  land,  how  much 
might  be  obtained  for  §  43.75  ? 

40.  If  9  hogsheads  of  sugar  cost  S  71.87,  what  cost  f  of  a 
hogshead  ? 

41.  Paid  $  4.56§  J  for  f  of  a  hogshead  of  sugar ;  what  ought 
to  be  given  for  9  hogsheads  ? 

42.  If  19  men  can  grade  a  certain  road  in  111  days,  how 
long  would  it  require  47  men  to  perform  the  same  labor  ? 

43.  When  47  men  can  grade  a  certain  road  in  44||  days, 
how  long  would  it  require  19  men  to  perform  the  same  labor  ? 

44.  If  T4T  of  a  ton  of  hay  cost  $  9.20,  what  cost  17  tons  ? 

45.  When  $  430.10  are  paid  for  17  tons  of  hay,  what  cost 
T4T  of  a  ton  ? 

46.  If  ft-  of  a  tub  of  butter  cost  $  7.15,  what  cost  7  tubs  ? 

47.  When  $  114.40  are  paid  for  7  tubs  of  butter,  what  cost 
-ft  of  a  tub  ? 

48.  If  a  horse  eat  19  f-  bushels  of  oats  in  87f-  days,  how  many 
will  7  horses  eat  in  60  days  ?  Ans.  93£  bushels. 

49.  Henry  Smith  can  reap  a  field  in  10  days,  by  laboring  8 
hours  a  day.  His  son  John  can  reap  the  same  field  in  9  days, 
by  laboring  12  hours  a  day.  How  long  would  it  take  both  to 
reap  the  field,  provided  they  labored  8  hours  a  day  ? 

Ans.  5£f  days. 


DECIMAL   FRACTIONS.  197 


DECIMAL   FRACTIONS. 

259.  A  Decimal  Fraction  is  a  fraction  whose  denom- 
inator is  some  power  often. 

It,  therefore,  originates  from  dividing  a  unit,  first  into  10 
equal  parts,  and  then  each  of  these  parts  into  10  other  equal 
parts,  and  so  on  indefinitely,  so  that  its  fractional  units  are 
tenths,  hundredths,  thousandths,  or  some  like  order  of  parts. 

260.  Decimal  fractions,  their  denominators  being  obvious, 
are  commonly  expressed  by  writing  the  numerator  only,  with 
the  decimal  point  (.)  before  it,  care  being  taken  to  put  a  cipher 
in  any  decimal  place  not  requiring  a  digit ;  thus,  . 

T9(5-  may  be  written  .9         and  be  read   9  tenths. 

TV^  "  .13  "  13  hundredths. 

T^ny  "  .005  "  5  thousandths. 

T#Hfo  "  -0105  "  105  ten-thousandths. 

261  •  By  examining  the  foregoing  fractions,  it  will  be  seen 
that,  — 

1.  The  denominator  of  a  decimal  fraction  is  1  with  as  many 
ciphers  annexed  as  the  numerator  has  places  of  figures, 

2.  In  writing  a  decimal  fraction  without  its  denominator, 
every  decimal  place  not  having  a  significant  figure  must  he 
filled  by  a  cipher. 

3.  The  first  figure  or  place  of  a  decimal  fraction  on  the 
right  of  the  decimal  point  is  tenths  ;  the  second,  hundredths  ; 
the  third,  thousandths  ;  the  fourth,  ten-thousandths  ;  fyc. 

4.  Each  figure  in  the  expression  of  decimal  fractions,  as  in 
whole  numbers,  represents  value,  according  to  its  distance  from 
the  place  of  units. 

262.  A  whole  number  and  a  decimal  fraction,  in  a  single 
expression,  constitute  a  mixed  number.  Thus,  17.63  is  a 
mixed  number,  and  is  read  seventeen,  and  decimal  sixty-three 
hundredths ;  150.302,  read  one  hundred  and  fifty,  and  decimal 
three  hundred  and  two  thousandths. 

Note.  —  For  the  sake  of  brevity,  especially  in  reading  mixed  numbers,  as  in 
the  instance  just  given,  a  decimal  fraction  is  commonly  called  simply  a  deci- 
mal. 

17* 


198  DECIMAL   FRACTIONS. 

263.  If  ciphers  are  placed  on  the  left  of  decimal  figures, 
between  them  and  the  decimal  point,  those  figures  change  their 
places,  each  cipher  removing  them  one  place  to  the  right,  and 
thus  diminishing  the  value  represented  tenfold.  Thus,  .9  =  t9q, 
but  .09  =  TJfo,  and  .009  —  TXfa. 

264.  If  ciphers  are  placed  on  the  right  of  decimal  figures, 
or  are  taken  away,  since  their  places  remain  the  same,  the 
value  represented  is  not  changed.     Thus,  .7  =  ^  and  .70  = 

265.  By  regarding  the  dollar  as  a  unit,  we  may  consider 
cents  and  mills  of  United  States  money  as  fractional  parts  of  a 
decimal  character.  Thus,  3  dollars  and  25  cents,  is  3  dollars 
and  25  hundredths  of  a  dollar,  or  $3.25;  also,  10  dollars  12 
cents  and  5  mills,  is  10  dollars  and  125  thousandths  of  a  dol- 
lar, or  $  10.125. 

NOTATION  AND  NUMERATION  OF  DECIMALS. 

266.  The  relation  of  decimals  to  whole  numbers  and  to 
each  other,  and  also  the  names  of  their  different  orders  and 
places,  are  shown  by  the  following 

Table. 

8  1  ill 


o 


O 

UJ 

G 

U 

O 

3 

T3 

a 

•8    1  4  I 

s    3    -i    k    i  J  *   I    3    ^    ■?    -2    ^    "2    .2 

o^s     §     a     s     o     s    Ja     S    ^    3     o    >2    fl 
HHKHPPHWHHW§HWpq 

654321.234567892 


0)         O        O         03         <X>         O         Q)  Cs(r><r><VOOOO<V 

ooooooo  ooooooooo 

co        e3      ^        o3        e3        c3        c3  C3c3c3s3c3cgccc3c5 

ooooooo  ooooooooo 

**   vi  **      i*      u      u      Ut  uuii*4t-'t-<*-*u£i 

O         O        O         O         ©         ©         ©  ©©©©©©©OO 

^           J_           ^            ^           *«           ^           £  ^^^^^Jh^*-!^ 

ooooooo  ooooooooo 

■5     rg    £     £     ^*     t>     -S  "S^^^^^rS^^ 

t~        q?        lO        rf        CO        <M        tH  TH<MCO-^iOCOt>-OOC5 


Whole  Numbers.  Decimals. 


DECIMAL   FRACTIONS. 


199 


Of  the  mixed  number  expressed  in  the  table,  the  part  on 
the  left  of  the  decimal  point  is  the  whole  number,  and  that  on 
the  right  the  decimal.  The  decimal  part  is  numerated  from 
the  left  to  the  right,  and  the  value  represented  is  expressed  in 
words  thus :  Two  hundred  thirty-four  million  five  hundred 
sixty-seven  thousand  eight  hundred  ninety-two  billionths.  And 
the  mixed  number  thus :  Seven  million  six  hundred  fifty-four 
thousand  three  hundred  twenty-one,  and  decimal  two  hundred 
thirty-four  million  five  hundred  sixty-seven  thousand  eight  hun- 
dred ninety-two  billionths. 

267.     From  the  table  we  deduce  the  following  rules : 

1.  Read  a  decimal  as  though  it  were  a  whole  number,  adding 
the  name  of  the  right-hand  order. 

2.  Write  a  decimal  as  though  it  were  a  whole  number,  sup- 
plying with  ciphers  such  places  as  have  no  significant  figures. 


Examples. 

Express  orally,  < 

[>r  write  in  words  the  followin 

g  numbers :  — 

1.         .056 

6. 

1.631 

11. 

1.000007 

2.         .1003 

7. 

48.07 

12. 

5.101016 

3.         .2786 

8. 

1.315 

13. 

1.000327 

4.         .16302 

9. 

5.6001 

14. 

0.000001 

o.         .97500 

10. 

87.0006 

15. 

16.000000007 

Express  in  the  c 

lecimal  form  by  figures  :  — 

16.         TV3o- 

17                  6 

18-        TirVW 

19. 
20. 
21. 

406 

To~(T(nro~ 
i 

TT5"0"0"T5"UT5" 

10  3  1 
Tff<J(5"G>(T(T 

22. 
23. 
24. 

7       17 

333^^^ 
iTotftjxnnr 

25.  Three  hundred  twenty-five,  and  seven  tenths. 

26.  Four  hundred  sixty-five,  and  fourteen  hundredths. 

27.  Ninety-three,  and  seven  hundredths. 

28.  Twenty-four,  and  nine  millionths. 

29.  Two  hundred  twenty-one,  and  nine  hundred-thousandths. 

30.  Forty-nine  thousand,  and  forty-nine  thousandths. 

31.  Seventy-nine  million  two  thousand,  and  one  hundred  five 
thousandths. 

32.  Sixty-nine  thousand  fifteen,  and  fifteen  hundred-thou- 
sandths. 


200  DECIMAL   FRACTIONS. 

33.  Eighty  thousand,  and  eighty-three  ten-thousandths. 

34.  Nine  billion  nineteen  thousand  nineteen,  and  nineteen 
hundredths. 

35.  Twenty-seven,  and  nine  hundred  twenty-seven  thou- 
sandths. 

36.  Forty-nine  trillion,  and  one  trillion th. 

37.  Twenty-one,  and  one  ten-thousandth. 

38.  Eighty-seven  thousand,  and  eighty-seven  millionths. 

39.  Ninety-nine  thousand  ninety-nine,  and  nine  thousand 
nine  billionths. 

40.  Seventeen,  and  one  hundred  seventeen  ten-thousandths. 

41.  Thirty-three,  and  thirty-three  hundredths. 

42.  Forty-seven  thousand,  and  twenty-nine  ten-millionths. 

43.  Fifteen,  and  four  thousand  seven  hundred-thousandths. 

44.  Eleven  thousand,  and  eleven  hundredths. 

45.  Seventeen,  and  eighty-one  quadrillionths. 

46.  Nine,  and  fifty-seven  trillionths. 

47.  Sixty-nine  thousand,  and  three  hundred  forty-nine  thou- 
sandths. 

268.  Decimals,  since  they  increase  from  right  to  left,  and 
decrease  from  left  to  right,  by  the  scale  of  ten,  as  do  simple 
whole  numbers,  may  be  added,  subtracted,  multiplied,  and 
divided  in  the  same  manner. 

ADDITION   OF  DECIMALS. 

269,  Ex.  1.  Add  together  23.61,  161.5,  2.6789,  and 
61.111.  Ans.  248.8999. 

OPERATION.  .^  . 

2  3  6  1  e  wnte  tne  numbers  so  that  figures  of  the 

-  £    "  k  same  decimal  place  shall  stand  in  the  same  col- 

o^  _  Q  a  unin,  and  then,  beginning  at  the  right  hand,  add 

2.6  7  8  J  them  as  whole  numbers  are  added,  and  place 

6  1.1  1  1  the  decimal  point  in  the  result  directly  under 

2  4  8.8  9  9  9       those  above* 

Rule.  —  Write  the  numbers  so  that  figures  of  the  same  decimal 
place  shall  stand  in  the  same  column. 

Add  as  in  whole  numbers,  and  point  off  in  the  sum,  from  the  right 
hand,  as  many  places  for  decimals  as  equal  the  greatest  number  of  deci- 
mal places  in  any  of  the  numbers  added. 


.DECIMAL   FRACTIONS.  201 

Proof.  —  The  proof  is  the  same  as  in  addition  of  whole 
numbers. 

Examples. 

2.  Add  together  the  following  numbers  :  81.61356,  6716.31, 
413.1678956,  35.14671,  3.1671,  314.6.    Ans.  7564.0052656. 

3.  "What  is  the  sum  of  the  following  numbers :  1121.6116, 
61.87,  46.67,  165.13,  676.167895?  Ans.  2071.449495. 

4.  Add  7.61,  637.1,  6516.14,  67.1234,  6.1234  together. 

Ans.  7234.0968. 

5.  Add  21.611,  6888.32,  3.6167  together. 

Ans.  6913.5477. 

6.  Add  together  $  15.06,  $  107.09,  $  1.625,  and  $  93.765. 

7.  I  have  bought  a  horse  for  $  137.50,  a  wagon  for  %  55.63, 
a  whip  for  $  1.375,  and  a  halter  for  %  0.87£ ;  what  did  they  all 
cost?  Ans.  $195.38. 

8.  What  is  the  sum  of  twenty-three  million  ten ;  one  thou- 
sand, and  five  hundred-thousandths ;  twenty-seven,  and  nineteen 
millionths  ;  seven,  and  five  tenths  ?     Ans.  23001044.500069. 

9.  Add  the  following  numbers:  fifty-nine,  and  fifty-nine 
thousandths ;  twenty-five  thousand,  and  twenty-five  ten-thou- 
sandths ;  five,  and  G.Ye  millionths ;  two  hundred  five,  and  five 
hundredths.  Ans.  25269.111505. 

10.  What  is  the  sum  of  the  following  numbers  :  twenty-five, 
and  seven  millionths ;  one  hundred  forty-five,  and  six  hundred 
forty-three  thousandths  ;  one  hundred  seventy-five,  and  eighty- 
nine  hundredths ;  seventeen,  and  three  hundred  forty-eight 
hundred-thousandths?  Ans.  363.536487. 

11.  A  farmer  has  sold  at  one  time  3  tons  and  75  hundredths 
of  a  ton  of  hay,  at  another  time  11  tons  and  7  tenths  of  a  ton, 
and  at  a  third  time  16  tons  and  125  thousandths  of  a  ton.  How 
much  has  he  sold  in  all  ?  Ans.  31.575. 

12.  Add  together  73  and  29  hundredths,  87  and  47  thou- 
sandths, 3005  and  116  ten-thousandths,  28  and  3  hundredths, 
29000  and  5  thousandths. 

13.  Add  two  hundred  nine  thousand  and  forty-six  millionths, 
ninety-eight  thousand  two  hundred  seven  and  fifteen  ten-thou- 
sandths, fifteen  and  eight  hundredths,  and  forty-nine  ten-thou- 
sandths, together.  Ans.  307222.086446. 


202  DECIMAL  FEACTIONS. 


SUBTRACTION  OF  DECIMALS. 

270.     Ex.  1.  From  61.9634  take  9.182.      Ans.  52.7814. 

operation.  Having  written  the  less  number  under  the  great- 

6  1.9  6  3  4  er,  so  that  figures  of  the  same  decimal  place  stand 

9.1  8  2  in  the  same  column,  we  subtract  as  in  whole  num- 

bers,  and  place  the  decimal  point  in  the  result,  as  in 

5  2.7  8  1  4  addition  of  decimals. 

Rule.  —  Write  the  less  number  under  the  greater,  so  that  figures  of 
the  same  decimal  place  shall  stand  in  the  same  column. 

Subtract  as  in  whole  numbers,  and  point  off  the  remainder  as  in 
addition  of  decimals. 

Proof.  —  The  proof  is  the  same  as  in  subtraction  of  whole 
numbers. 

Examples. 
2.  3.  4.  5. 

3  9.3  5.  6.1  41.7 

1.678  9  1.6  7  8  1.99999  21.9767 


3  7.62  11  3.3  2  2  4.10001  19.7233 

6.  From  29.167  take  19.66711.  Ans.  9.49989. 

7.  From  91.61  take  2.6671.  Ans.  88.9429. 

8.  From  96.71  take  96.709. 

9.  Take  twenty-seven  and  twenty-eight  thousandths  from 
ninety-seven  and  seven  tenths.  Ans.  70.672. 

10.  Take  one  hundred  fifteen  and  seven  hundredths  from 
three  hundred  fifteen  and  twenty-seven  ten-thousandths. 

Ans.  199.9327. 

11.  From  twenty-nine  million  four  thousand  and  five  take 
twenty-nine  thousand,  and  three  hundred  forty-nine  thousand 
two  hundred,  and  twenty-four  hundred-thousandths. 

Ans.  28625804.99976. 

12.  From  one  million  take  one  millionth. 

Ans.  999999.999999. 

13.  From  $  19  take  $  1.375.  Ans.  $  17.625. 

14.  A  merchant  bought  flour  to  the  amount  of  $316.87£, 
and  sold  it  for  $  400  ;  how  much  did  he  gain  by  the  sale  ? 

15.  From  19  million  take  19  billionths. 

Ans.  18999999.999999981. 


DECIMAL   FRACTIONS.  203 

16.  Charles  Washburne  has  in  one  farm  93.45  acres,  in 
another  124  acres,  in  a  third  244.285  acres,  and  in  wood-lots 
216.136  acres;  how  many  acres  more  would  he  require  to 
have  exactly  1000  acres  ? 

MULTIPLICATION  OF   DECIMALS. 

271.     Ex.  1.  Multiply  76.81  by  3.2.  Ans.  245.792. 

operation.  we  multiply  as  in  whole  numbers,  and  point  off 

7  6.8  1  on  the  right  of  the  product  as  many  figures  for  deci- 

3.2  mals  as  there  are  decimal  figures  in  the  multiplicand 

^  and  multiplier  counted  together.     The  reason  for 

1  0  6  b  L  pointing  off  the  decimals  in  the  product,  as  in  the 

2  3  0  4  3  operation,  will  be  seen,  if  we  convert  the  multipli- 

2  4  5  7  9  2  cand  and  multiplier  into  common  fractions,  and  mul- 

tiply  them  together.     Thus,  76.81  =  76^  =  ^1 ; 

PT1/|   Q9  Q2 82  Thpn    .7  68  1    y    3  2.  245792  94.5  7  9_2__  _ 

ana  6.L  —  6^-§  —  t-q.       j_nen  ^oo    *  .To  —    1000    —  Z40ToFo  — 
245.792,  Ans.,  the  same  as  m  the  operation. 

2.  Multiply  .1234  by  .0046.  Ans.  .00056764. 

operation.  Since  the  number  of  figures  in  the  product 

.12  3  4      is  not  equal  to  the  number  of  decimals  in  the 

.0046      multiplicand  and  multiplier,  we  supply  the  de- 

ficiency  by  placing  ciphers  on  the  left  hand. 

7  4  0  4      ■j'jjg  reason  of  this  process  will  appear,  if  we 

4  9  3  6         perform  the  operation  thus  :  .1234  =  ^AVo  ; 

00056764     an*  -0??6  =  to4Ao.     Then^AX^Aor 
.v  v  v  o  o  i  o  *  56764qo  =  .00056764,  Ans.,  the  same  as  in 

the  operation. 

Rule.  —  Multiply  as  in  whole  numbers,  and  point  off  as  many 
figures  for  decimals,  in  the  product,  as  there  are  decimal  figures  in  the 
multiplicand  and  multiplier. 

If  there  be  not  so  many  figures  in  the  product  as  there  are  decimal 
figures  in  the  multiplicand  and  multiplier,  supply  the  deficiency  by  pre- 
fixing ciphers. 

Proof,  —  The  proof  is  the  same  as  in  multiplication  of  whole 
numbers. 

Examples. 

3.  Multiply  61.76  by  .0071.  Ans.  .438496. 

4.  Multiply  .0716  by  1.326.  Ans.  .0949416. 

5.  Multiply  .61001  by  .061. 

6.  Multiply  71.61  by  365.  Ans.  26137.65. 

7.  Multiply  .1234  by  1234.  Ans.  152.2756. 


204  DECIMAL   FRACTIONS. 

8.  Multiply  6.711  by  6543.  Ans.  43910.073. 

9.  Multiply  .0009  by  .0009.  Ans.  .00000081. 

10.  What  is  the  product  of  one  thousand  and  twenty-five, 
multiplied  by  three  hundred  and  twenty-seven  ten-thousandths  ? 

Ans.  33.5175. 

11.  What  is  the  product  of  seventy-eight  million  two  hun- 
dred five  thousand  and  two,  multiplied  by  fifty-three  hun- 
dredths ?  Ans.  41448651.06. 

12.  Multiply  one  hundred  and  fifty-three  thousandths  by  one 
hundred  twenty -nine  millionths.  Ans.  .000019737. 

13.  What  will  26.7  yards  of  cloth  cost,  at  $5.75  a  yard? 

Ans.  $153,525. 

14.  What  will  14.75  bushels  of  wheat  cost,  at  $  1.25  a 
bushel  ?  Ans.  $  18.4375. 

15.  What  will  375.6  pounds  of  sugar  cost,  at  $0,125  per 
pound  ? 

16.  What  will  26.58  cords  of  wood  cost,  at  $5,625  a 
cord?  Ans.  $149.512£. 

17.  What  will  28.75  tons  of  potash  cost,  at  $  125.78  per 
ton?  Ans.  $3616.175. 

CONTRACTIONS  IN  MULTIPLICATION  OF  DECIMALS. 

272.  To  multiply  a  decimal  by  10,  100,  1000,  &c.  Re- 
move  the  decimal  point  as  many  places  to  the  right  as  there  are 
ciphers  in  the  multiplier,  annexing  ciphers  if  required.  Thus, 
1.25  X  10  =  12.5  ;  and  1.6  X  100  =  160. 

Examples. 

1.  Multiply  131.634  by  1000.  Ans.  131634. 

2.  Multiply  3478.9  by  100.  Ans.  347890. 

3.  Multiply  one  thousandth  by  one  thousand. 

4.  What  is  the  profit  on  one  million  yards  of  cotton  cloth,  at 
$  0.007  per  yard.  Ans.  $  7000. 

273.  When  it  is  not  necessary  that  all  the  decimal  places 
of  the  product  should  be  retained,  tedious  multiplications  may 
often  be  obviated,  by  contracting  the  work  as.  follows  :  — 


DECIMAL   FRACTIONS. 


205 


Write  the  units'  place  of  the  multiplier  under  that  figure  of 
the  multiplicand  whose  place  it  is  proposed  to  retain  in  the  pro- 
duct, and  dispose  of  all  the  other  figures  of  the  multiplier  in 
an  order  contrary  to  the  usual  one.  Then,  in  multiplying, 
begin,  for  each  partial  product,  with  that  figure  of  the  multipli- 
cand which  stands  above  the  multiplying  figure,  observing  to  add 
to  the  product  the  number  nearest  to  that  which  would  have  been 
carried  if  the  places  at  the  right  had  not  been  rejected.  Write 
down  the  several  partial  products,  so  that  the  right-hand  figure 
of  each  shall  be  in  the  same  column,  and  their  sum  will  be  the 
product  required. 

Examples. 

1.  Multiply  3.141592  by  52.7438,  retaining  only  four  places 
for  decimals  in  the  product.  Ans.  165.6995 


FIRST    OPERATION. 


3.1  4  1  5  9  2  ==  Multiplicand 

8  3  4  7.2  5     =  Multiplier  reversed. 

=  Product  by  5,  -|~  1 
=  Product  by  2,  +  2 
=  Product  by  7,  -f-  4 
=  Product  by  4,  --  1 
=  Product  by  3,  +  1 
=  Product  by  8,  -f-  1 

1  6  5.6  9  9  5     =  Product  sought. 


1570796 

62832 

21991 

1257 

94 

25 


SECOND    OPERATION. 

3.1  4  1  5  9  2 

5  2.7  4  3  8 

25113  27  3  6 

9424776 
1256  6368 


21991 

62831 

1570796 


16  5.69  95 


144 

84 
0 


001296 


By  comparison  of  the  two  methods  of  solution,  it  will  be  seen  that 
the  common  one,  as  shown  in  the  second  operation,  gives  ten  places 
of  decimals,  or  six  more  than  are  required  by  the  question,  thus  ren- 
dering unnecessary  the  several  figures  on  the  right  of  the  vertical 
line.  By  the  contrasted  way,  the  multiplier,  for  convenience,  has  its 
figures  reversed,  or  placed  contrary  to  the  usual  order,  so  that  the 
product  of  each  figure  by  the  one  of  the  multiplicand  above  it,  must 
be  of  the  order  of  ten-thousandths.  The  first  figure,  at  the  right,  of 
each  partial  product,  being  of  the  order  of  ten-thousandths,  is  writ- 
ten in  the  same  column.  To  the  product  by  5  we  add  1,  since,  if 
the  2  in  the  multiplicand  had  not  been  rejected,  there  would  have 
been  1  to  carry  to  the  product  of  the  9  by  the  5  ;  to  the  product  by 
2  we  add  2,  since  the  product  of  the  rejected  figures,  92,  by  2, 
approximates  to  2  hundred,  which  would  require  2  to  be  carried ; 
to  the  product  by  7  we  add  4,  since  the  product  of  the  two  re- 
jected figures,  59,  by  7,  would  require  4  to  be  carried ;  to  the  pro- 
N    18 


206  DECIMAL  FRACTIONS. 

duct  by  4  we  add  1,  since  the  product  of  the  two  rejected  figures, 
15,  by  4,  approximates  to  1  hundred,  which  would  require  1  to  be 
carried ;  and  so  on,  it  being  sufficient  to  increase  the  partial  product 
only  by  such  a  number  as  approximates  most  nearly  to  that  which 
would  have  been  carried,  provided  the  two  rejected  figures  next  to 
the  figure  of  the  multiplicand  had  been  retained. 

2.  Multiply  325.701428  by  .7218393,  retaining  only  three 
places  of  decimals  in  the  product.  Ans.  235.104. 

3.  Multiply  56.7534916   by  5.376928,  retaining  only  five 
places  of  decimals  in  the  product. 

4.  Multiply  843.7527  by  8634.175,  retaining  only  the  in- 
tegers in  the  product.  Ans.  7285109. 

DIVISION  OF  DECIMALS. 

274.     Ex.  1.    Divide  1.728  by  1.2.  Ans.  1.44. 

operation.  "We  divide  as  in  whole  numbers, 

1.2  )  1.7  2  8  (  1.4  4  Ans.      and,  since  the  divisor  and  quotient 

1  2  are  the  two  factors,  which,  being  mul 


■7. 


tiplied  together,  produce  the  dividend, 

0  ^  we  point  olf  two  decimal  figures  in  the 

4  8  quotient,  to  make  the  number  in  the 

a  o  two  factors  equal  to  the  number  in 

the  product  or  dividend. 

The   reason   for   pointing  off  will 
also  be  seen  by  performing  the  ex- 
ample with  the  decimals  in  the  form  of  common  fractions.     Thus, 
1.728  =  1TV&  =  «»  ;  and  1 .2  =  1  ft  =  }§ .     Then  ft ft  -  ft 
=  tttt  X  «  -  \UU  =  m=  *AV  =  I-**,  Ans.  as  before. 

2.  Divide  36.6947  by  589.  Ans.  .0623. 

operation.  .  "We   divide    as    in   whole 

589  )  3  6. 6947  (.0  623  Ans.      numbers,  and  since  we  have 

3  5  3  4  .  but  three  figures  in  the  quo- 

—  tient,  we  place  a  cipher  be- 

1  "  °  ^  fore  them,  and  thus  make  the 

117  8  decimal  places  in  the  divisor 

^  rr  n  rj  and  quotient  equal  to  those 

.j  _  c  „  of  the  dividend. 

The   reason  for  prefixing 

the  cipher  will  appear  more 

obvious  by  solving  the  example  in  the  form  of  common  fractions. 

Thus,  86.6947  =  86^^  =  V&W  <  and   589  —  *P-      Then 

366947      »      £8.2.   __    366047     y      .1       3  6  6  9  4  7      623       Ofi^S 

100  00     7~        1       —    "TWFiT    *    X8^  5~8"  9  001  IT   ~   TOOOlT   —   -^-3, 

Ans.  as  before.     Hence  the  following 


DECIMAL    FRACTIONS. 


207 


Rule.  — Divide  as  in  ivhole  numbers,  and  point  off  as  many  figures 
in  the  quotient  as  the  number  of  decimal  places  in  the  dividend  exceeds 
the  number  in  the  divisor ;  but  if  there  are  not  as  many,  supply  the 
deficiency  by  prefixing  ciphers. 

Note  1.  —  When  the  decimal  places  in  the  divisor  exceed  those  in  the  divi- 
dend, make  them  equal  by  annexing  ciphers  to  the  dividend,  and  the  quotient 
will  be  a  whole  number. 

Note  2.  —  When  there  is  a  remainder  after  dividing  the  dividend,  ciphers 
may  be  annexed,  and  the  division  continued ;  the  ciphers  thus  annexed  being 
regarded  as  decimals  of  the  dividend ;  and  to  indicate  in  any  case  that  the 
division  does  not  terminate,  the  sign  plus  (  +  )  can  be  used. 

Proof.  —  The  proof  is  the  same  as  in  division  of  whole 
numbers. 

Examples. 


Ans.  13.861+. 

Ans.  749.084. 

Ans.  3124.3. 


3.  Divide  780.516  by  2.43.  Ans.  321.2. 

4.  Divide  7.25406  by  9.57.  Ans.  .758. 

5.  Divide  .21318  by  .38. 

6.  Divide  7.2091365  by  .5201. 

7.  Divide  56.8554756  by  .0759. 

8.  Divide  119109094.835  by  38123.45. 

9.  Divide  1191090.94835  by  3812345. 

10.  Divide  11910909483.5  by  38.12345. 

11.  Divide  11.9109094835  by  381234.5. 

12.  Divide  1191.09094835  by  3.812345. 

13.  Divide  11910909483.5  by  .3812345. 

14.  Divide  1.19109094835  by  3.812345. 

15.  Divide  .119109094835  by  .3812345. 

16.  Divide  30614.4  by  .9567.  Ans.  32000. 

17.  Divide  .306144  by  9567.  Ans.  .000032. 

18.  Divide  four  thousand  three  hundred  twenty-two  and  four 
thousand  five  hundred  seventy-three  ten-thousandths  by  eight 
thousand  and  nine  thousandths.  Ans.  .5403—)—. 

19.  How  many  yards  of  calico  at  $0.0775  per  yard  can  be 
purchased  for  $  10.85  ? 

20.  What  costs  1  acre  of  woodland  when  19.65  acres  are 
sold  for  $  982.50  ?  Ans.  $50. 

21.  Divide  three  hundred  twenty-three  thousand  seven  hun- 
dred sixty-five  by  five  millionths.  Ans.  64753000000. 


208  DECIMAL   FRACTIONS. 


CONTRACTIONS  IN  DIVISION  OF  DECIMALS. 

275,  To  divide  a  decimal  by  10,  100,  1000,  &c. 

Remove  the  decimal  point  as  many  places  to  the  left  as  there 
are  ciphers  in  the  divisor,  and  if  there  be  not  figures  enough  in 
the  number,  prefix  ciphers.  Thus,  2.15  -f-  10  =  .215 ;  and 
1.9  -T-  100  —  .019. 

Examples. 

1.  Divide  31.675  by  10. 

2.  Divide  916.05  by  100. 

3.  Divide  7.0461  by  100000. 

4.  Divide  70.461  by  100000. 

5.  Divide  704.61  by  100000. 

6.  Divide  7046.1  by  100000. 

7.  Divide  70460  by  100000. 

8.  Divide  .70460  by  100000. 

9.  Divide  196.5  by  1000000.  Ans.  .0001965. 

10.  If  $3500  are  paid  for  1000  yards  of  broadcloth,  what  is 
it  a  yard  ?  Ans.  $  3.50. 

11.  When  $  1025  are  paid  for  40  boxes  of  sugar,  each  con- 
taining 250  pounds,  what  is  the  cost  of  1  pound  ? 

Ans.  $0.10£. 

276.  When  the  divisor  contains  many  decimal  places,  and 
only  a  certain  number  of  decimals  are  required  to  be  retained 
in  the  quotient,  the  work  may  be  contracted  as  follows  :  — 

First  consider  how  many  figures,  in  all,  it  is  necessary  for  the 
quotient  to  contain.  Then,  by  using  the  same  number  of  figures 
from  the  left  of  the  divisor,  find  the  first  figure  of  the  quotient, 
and,  instead  of  bringing  down  a  new  figure  from  the  dividend, 
or  annexing  a  cipher  to  the  remainder,  reject  a  figure  on  the 
right  of  the  divisor  at  each  successive  division,  and  make  the 
other  figures  a  divisor.  In  multiplying  such  a  divisor  by  the 
quotient  figure,  observe  to  add  to  the  product  the  number  nearest 
to  that  which  would  have  been  carried  if  no  figures  had  been 
rejected. 


DECIMAL   FRACTIONS. 


209 


Examples. 

1.  Divide  695.57270875  by  52.35775,  and  retain  in  the  quo- 
tient three  places  of  decimals. 


FIRST  OPERATION. 


5  2.3  5  7  7  5  )  6  9  5.5  7  2  7  0  8  7  5  (  1  3.2  8  5. 
52358  =  product  by  1,  +  1. 

17  19  9 

15707  =  product  by  3,  +  2. 

1492 

10  4  7:=  product  by  2,  -f-  1. 

product  by  8,  -\-  3. 

~26 
2  6=  product  by  5,  -f-  1- 

SECOND    OPERATION. 

5  2.3  5775)695.5  7270875(13.2  8  5. 
5  2  35775 


445 
419 


17199 
15707 


1492 
1047 


520 
325 


Ans.  13.285. 

By  inspection, 
it  is  evident  that 
the  first  quotient 
figure  will  be 
of  the  order  of 
tens,  and  there- 
fore the  quotient 
will  contain  two 
places  of  whole 
numbers ;  and  as 
there  are  to  be 
three  places  of 
decimals,  it  must 
contain  five  fig- 
ures. Hence,  we 
divide  at  first  by 
five  figures  of 
the  given  di- 
visor, counting 
them  from  the 
left  toward  the 
right,  thus  using 
the  52.357  and 
rejecting  the 
figures,  75,  on 
the  right.  In 
multiplying  each 
contracted  divi- 
sor by  its  quo- 
tient figure  we 
increase  the  pro- 
duct by  having  regard  to  rejected  figures,  as  in  contracted  multiplica- 
tion of  decimals  (Art.  273). 

The  nature  and  extent  of  the  contraction  will  be  seen  by  compari- 
son with  the  common  method  as  shown  in  the  second  operation,  in 
which  the  vertical  line  cuts  off  the  figures  not  required. 

Note.  —  When  the  given  divisor  does  not  contain  as  many  figures  as  are  re- 
quired in  the  quotient,  we  must  begin  the  division  in  the  usual  way,  and  con- 
tinue till  the  deficiency  is  made  up,  after  which  begin  the  contraction. 

2.  Divide  4327.56284563  by  873.469,  and  retain  five  decimal 
places  in  the  quotient.  Ans.  4.95445. 

3.  Divide  252070.520751  by  591.57,  and  terminate  the  opera- 
tion with  four  decimal  places  in  the  quotient.     Ans.  426.1043. 

18* 


1958 
1550 


04087 
86200 

178875 
178875 


210  DECIMAL   FEACTIONS. 

4.  Divide  70.23  by  7.9863,  and  retain  in  the  answer  four 
decimals. 

5.  Divide  12193263.1112635269  by  123456789,  and  let  the 
quotient  contain  as  many  decimal  places,  plus  one,  as  there  will 
be  integers  in  it.  Ans.  9876.54321. 

EEDUCTION  OF  DECIMALS. 

277.     To  reduce  a  decimal  to  a  common  fraction. 
Ex.  1.     Reduce  .125  to  its  equivalent  common  fraction. 


Ans.  £. 


OPERATION. 


•125  =  rifift-  =  jflfr  =  A  =  i  An8' 


Erasing  the  decimal  point  and  supplying  the  denominator,  which 
is  understood,  we  have  ^(HF*  which  reduced  to  its  lowest  terms  equals 
J,  the  answer  required. 

Rule.  —  Erase  the  decimal  point,  and  write  under  the  numerator  its 
decimal  denominator,  and  reduce  the  fraction  to  its  lowest  terms. 

Examples. 

2.  Reduce  .875  to  a  common  fraction.  Ans.  £. 

3.  Reduce  .9375  to  a  common  fraction.  Ans.  \% . 

4.  What  common  fraction  is  equivalent  to  .08125? 

Ans.  t^j. 

5.  Change  .00075  to  the  form  of  a  common  fraction. 

6.  Express  31.75  by  an  integer  and  a  common  fraction. 

Ans.  31£. 

7.  Express  96.024  by  an  integer  and  a  common  fraction. 

Ans.  96T^. 

8.  Express  163.04  by  an  integer  and  a  common  fraction. 

9.  Express  1001.4375  by  an  integer  and  a  common  frac- 
tion. Ans.  1001^. 

10.  Express  1457.222  by  an  integer  and  a  common  fraction. 

11.  Express  19678.36  by  an  integer  and  a  common  fraction. 

12.  Express  9163.8755  by  an  integer  and  a  common  frac- 
tion. Ans.  9163£$££. 

278 •     To  reduce  a  common  fraction  to  a  decimal. 

Ex.  1.     Reduce  &  to  a  decimal.  Ans.  .375. 


DECIMAL    FRACTIONS. 


211 


OPERATION. 


Since  we  cannot  divide  the  numer- 
ator, 3,  by  8,  we  reduce  it  to  tenths 
by  annexing  a  cipher,  and  then  di- 
viding, we  obtain  3  tenths  and  a  re- 
mainder of  6  tenths.  Reducing  this 
remainder  to  hundredths  by  annexing 
a  cipher,  and  dividing,  we  obtain  7 
hundredths  and  a  remainder  of  4 
hundredths ;  which  being  reduced  to 
thousandths  by  annexing  a  cipher, 
and  then  divided,  gives  a  quotient  of 
5  thousandths.  The  sum  of  the  sev- 
eral quotients,  .375,  is  the  answer. 

To  prove  that  .375  is  equal  to  f, 
we  change  it  to  the  form  of  a  com- 
mon fraction,  by  writing  its  denomi- 
nator, and  reducing  it  to  its  lowest  terms.     Thus,  .375  =  -f^-fa  =  f. 


8  )  3.0  (  3  tenths. 
24 

8)60(7  hundredths. 
5_6 

8)40(5  thousandths. 
4j0 

Ans.  .375. 

Or  thus : 

8  )  3.0  0  0 

.3  7  5  Ans. 


Rule.  —  Annex  ciphers  to  the  numerator,  and  divide  by  the  denomi- 
nator. Point  off  in  the  quotient  as  many  decimal  places  as  there  have 
been  ciphers  annexed. 

Note.  —  It  is  not  usually  necessary  that  the  decimals  should  be  carried  to 
more  than  six  places.  When  a  decimal  does  not  terminate,  the  sign  plus  (+)  is 
generally  annexed.  Thus,  in  the  expression  .333+,  the  sign  annexed  indicates 
that  the  division  could  be  carried  further. 


Examples. 

2.  Reduce  ■§■  to  a  decimal.  Ans.  .625. 

3.  Reduce  J  to  a  decimal. 

4.  Change  A-  to  a  decimal.  Ans.  .09375. 

5.  Change  T^  to  a  decimal.  Ans.  .076923-f-. 

6.  Reduce  19|-  to  an  equivalent  decimal  expression. 

7.  Reduce  $  31 5  J  to  an  equivalent  decimal  expression. 

Ans.  $315,875. 

8.  Reduce  $  1163f  to  an  equivalent  decimal  expression. 

Ans.  1163.75. 

Note.  —  A  decimal  with  a  common  fraction  annexed  constitutes  what  is 
called  a  complex  decimal;  as,  .87  j,  .311,  and  .182.  In  such  expressions,  instead 
of  the  common  fraction,  its  equivalent  decimal,  with  the  decimal  point  omitted, 
may  be  substituted.    Thus,  Afa  =  .404. 

9.  Reduce  .62^-  to  a  simple  decimal.  Ans.  .625. 

10.  Reduce  .37-^  to  a  simple  decimal.  Ans.  .370625. 

11.  Reduce  $  4.31  J-  to  a  simple  decimal  expression. 

Ans.  $  4.3125. 


212  DECIMAL   FKACTIONS. 

12.  Reduce  $  60.18J  to  a  simple  decimal  expression. 

Ans.  $  60.1875. 

13.  What  decimal  expression  is  equivalent  to  £  of  —  of  2.04? 

o 

14.  What  decimal  expression  is  equivalent  to  2^-,  -J-  0.37 J, 
+  £  of  $  of  4,  —  1.05  ?  Ans.  2.9875. 

279.  To  reduce  a  simple  or  compound  number  to  a  deci- 
mal of  a  higher  denomination. 

Ex.  1.     Reduce  15s.  9d.  3far.  to  the  decimal  of  a  £. 

Ans.  .790625. 

operation.  "VVe   commence  with  the  3far.,  which 

4     3.0  0  far.      we  reduce  to   hundredths  by  annexing 

1  9      9  7  5  0  0       1  two  °ipners5    and  then,  to  reduce  these 

to  the  decimal  of  a  penny,  we  divide  by 

2  0  1  5.8  12  5  0s.         4?    since    there  will   be  |  as  many  hun- 

dredths  of  a  penny  as  of  a  farthing,  and 

.790625  £.        obtain  .75d.     Annexing  this  to  the  9d., 

we  divide  by  12,  since  there  will  be  -fa  as 

many  shillings  as  pence;    and  then,  the  15s.  and  this  quotient  by 

20,  since  there  will  be  -£$  as  many  pounds  as  shillings,  and  obtain 

.790625£.  for  the  answer.     Hence  the  following 

Rule.  —  Divide  the  lowest  denomination,  annexing  ciphers  if  neces- 
sary, by  that  number  which  will  reduce  it  to  one  of  the  next  higher  de- 
nomination. Then  divide  as  before,  and  so  continue  dividing  till  the 
decimal  is  of  the  denomination  required. 

Note  1.  —  The  given  numbers  may  also  be  first  reduced  to  a  common  frac- 
tion (Art.  256),  and  then  the  fraction  changed  to  a  decimal.  Thus,  if  it  be 
required  to  reduce  15s.  6d.  to  a  decimal  o£a£.:  15s.  6d.  =  186d. ;  1£.  =  280d. ; 
Mo  £'  =  lo  £'  =  *775  £'  Answer. 

Note  2.  —  Shillings,  pence,  and  farthings  may  be  readily  reduced  to  a  deci- 
mal of  three  places,  by  inspection,  thus  :  Call  half  of  the  greatest  even  number 
of  shillings  tenths,  and,  if  there  be  an  odd  shilling,  call  it  5  hundredths;  re- 
duce the  pence  and  farthings  to  farthings,  and  increase  them  by  1,  if  they  amount 
to  24  or  more,  for  thousandths.  Thus,  if  it  be  required  to  reduce,  by  in- 
spection, 19s.  lOd.  2far.  to  the  decimal  of  a  £.;  half  of  18s.  =  9s.,  which  de- 
note a  value  of  .9£. ;  the  Is.  denotes  a  value  of  .05<£. ;  and  lOd.  2far.  =  42far., 
-which  increased  by  lfar.  =  43far.,  which  denote  a  value  of  .043£. ;  .9£.  + 
.05£.  +  .043£.  =  .993£.  Answer. 

The  reason  for  this  process  is,  that  2s.  equal  a  tenth  of  a  £. ;  1  shilling  equals 
5  hundredths  of  a  £.,  and  1  farthing  equals  §£5£.,  or  so  nearly  a  thousandth  of 
a  £.  that  24  farthings  exactly  equal  25  thousandths  of  a  £. ;  and  therefore 
farthings  require  to  be  increased  only  by  1  when  they  amount  to  24  or  more, 
to  denote  with  sufficient  accuracy  their  value  in  thousandths  of  a  £. 


DECIMAL    FRACTIONS.  213 

Examples. 

2.  Eeduce  9s.  to  the  fraction  of  a  pound.  Ans.  .45. 

3.  Reduce  15cwt.  3qr.  141b.  to  the  decimal  of  a  ton. 

4.  Eeduce  2qr.  211b.  8oz.  12dr.  to  the  decimal  of  a  cwt. 

Ans.  .71546875. 

5.  Eeduce  lqr.  3na.  to  the  decimal  of  a  yard. 

Ans.  .4375. 

6.  Eeduce  5fur.  35rd.  2yd.  2ft.  9in.  to  the  decimal  of  a  mile. 

Ans.  .73603219-f. 

7.  Eeduce  3gal.  2qt.  lpt.  of  wine  to  the  decimal  of  a  hogs- 
head. Ans.  .0575396+. 

8.  Eeduce  lpt.  to  the  decimal  of  a  bushel.     Ans.  .015625. 

9.  Eeduce  2E.  16p.  to  the  decimal  of  an  acre.        Ans.  .6. 

10.  Eeduce  175  cubic  feet  to  the  decimal  of  a  ton  of  timber. 

Ans.  4.375. 

11.  Eeduce  3.755  pecks  to  the  decimal  of  a  bushel. 

Ans.  .93875. 

12.  What  decimal  part  of  a  degree  is  25'  34".6? 

13.  Eeduce  12T.  3cwt.  2qr.  201b.  to  hundred-weight  and  the 
decimal  of  a  hundred- weight.  Ans.  243.7. 

14.  Eeduce  2hhd.  30gal.  2qt.  lj-pt.  to  gallons  and  the  decimal 
of  a  gallon.  Ans.  156.6875. 

15.  Eeduce  to  the  decimal  of  a  pound,  19s.  llfd.,  16s.  9£d., 
and  17s.  5|-d.,  and  find  their  sum.  Ans.  2.71041 6-f-. 

280.     To  find  the  value  of  a  decimal  in  whole  numbers  of 
lower  denominations. 

Ex.  1.     What  is  the  value  of  .790625  £.  ? 

Ans.  15s.  9d.  3far. 

operation.  There  will  be  20  times  as  many  mil- 

.79062  5£.  lionths  of  a  shilling  as  of  a  pound  ;  there- 

2  0  fore,  we  multiply  the  decimal,  .790625, 

a  by  20,  and  reduce  the  improper  fraction 

1  o.o  1  2  o  0  Os.  t0  a  mixed  number  by  pointing  off  six 

1  2  figures  on  the  right,  which  is  dividing  by 

9  7  5  0  0  0  Od  *ts  denominator,  1000000.     The  figures 

'  on  the  left  of  the  point  are  shillings,  and 

those  on  the  right,  the  decimal  of  a  shilling. 

3.0  0  0  0  0  Ofar.  ^n*s  decimal  of  a  shilling  we  multiply  by 

A         i  *     o  i    q**  *2'  anc*'  Pomtin»  off  as  before,  obtain  9d., 

Ans.  los.  Jd.  otar.  an(j  a  decimal  of  a  penny.     The  decimal 


214  MISCELLANEOUS   EXAMPLES. 

of  a  penny  we  multiply  by  4,  and  pointing  off  have  3  farthings,  which, 
taken  with  the  other  denominations  obtained,  gives  15s.  9d.  3far.  for 
the  answer. 

Rule.  —  Multiply  the  decimal  by  that  number  which  icill  reduce  it  to 
the  next  lower  denomination,  and  point  off  as  in  multiplication  of  deci- 
mals. 

Then,  multiply  the  decimal  part  of  the  product,  and  point  off  as  be- 
fore. So  continue  till  the  decimal  is  reduced  to  the  denominations 
required. 

The  several  whole  numbers  of  the  successive  products  ivill  be  the 
answer. 

Note.  —  When  there  is  a  decimal  in  the  last  product,  it  may  be  changed  to 
a  common  fraction. 

Examples. 

2.  What  is  the  value  of  .625  of  a  shilling  ?  Ans.  7Jd. 

3.  What  is  the  value  of  .6725  of  a  cwt.  ? 

Ans.  2qr.  171b.  4oz. 

4.  What  is  the  value  of  .9375  of  a  yard  ? 

5.  What  is  the  value  of  .7895  of  a  mile  ? 

Ans.  6fur.  12rd.  10ft.  6Jfin. 

6.  What  is  the  value  of  .9378  of  an  acre  ? 

Ans.  3R.  30p.  13ft.  9T9^in. 

7.  Reduce  .5615  of  a  hogshead  of  wine  to  its  value  in  gal- 
lons, &c.  Ans.  35gal.  lqt.  Opt.  3||fgi. 

8.  Reduce  .367  of  a  year  to  its  value  in  days,  &c. 

Ans.  134d.  lh.  7m.  19£sec. 

9.  What  is  the  value  of  .6923828125  of  a  cwt.  ? 

Ans.  2qr.  191b.  3oz.  13d.. 

10.  What  is  the  value  of  .015625  of  a  bushel  ? 

11.  What  is  the  value  of  .55  of  an  ell  English  ? 

Ans.  2qr.  3na. 

12.  What  is  the  value  of  .6  of  an  acre  ?        Ans.  2R.  16p. 


MISCELLANEOUS  EXAMPLES. 

1.  What  is  the  value  of  7cwt.  2qr.  181b.  of  sugar,  at  $  11.75 
per  cwt.  ?  Ans.  $  90.24. 

2.  What  cost  19cwt.  3qr.  141b.  of  iron,  at  $  9.25  per  cwt.  ? 


MISCELLANEOUS    EXAMPLES.  215 

3.  What  cost  39A.  2R.  15p.  of  land,  at  $87,375  per 
acre  ?  Ans.  $  3459.503|f . 

4.  What  would  be  the  expense  of  making  a  turnpike  87m. 
3fur.  15rd.,  at  $  578.75  per  mile  ?  Ans.  $  50595.41^¥. 

5.  What  is  the  cost  of  a  board  18ft.  9in.  long,  and  2ft.  3£in. 
wide,  at  $  .053  per  foot  ?  Ans.  $  2.277^. 

6.  Goliath  of  Gath  was  6 \  cubits  high ;  what  was  his  height 
in  het,  the  cubit  being  1ft.  7.168in.  ?         Ans.  10ft.  4.592in. 

7.  If  a  man  travel  4.316  miles  in  an  hour,  how  long  would 
he  be  in  travelling  from  Bradford  to  Boston,  the  distance  being 
29J  miles  ?  Ans.  6h.  50m.  6sec.-f 

8.  What  is  the  cost  of  5yd.  lqr.  2na.  of  broadcloth,  at  $  5.62  J 
per  yard  ?  Ans.  $  30.234f . 

9.  Bought  17  bags  of  hops,  each  weighing  4cwt.  3qr.  71b., 
at  $  5.87£  per  cwt. ;  what  was  the  cost  ? 

10.  Purchased  a  farm,  containing  176A.  3R.  25rd.,  at 
$  75.37£  per  acre ;  what  did  it  cost  ?      Ans.  $  13334.308  J£. 

11.  What  cost  17625  feet  of  boards,  at  $12.75  per  thou- 
sand ?  Ans.  $  224.718J . 

12.  How  many  square  feet  in  a  floor  19ft.  3in.  long,  and 
15ft.  9in.  wide  ?  Ans.  303ft.  27in. 

13.  How  many  square  yards  of  paper  will  it  take  to  cover  a 
room  14ft.  6in.  long,  12ft.  6in.  wide,  and  8ft.  9in.  high? 

14.  How  many  solid  feet  in  a  pile  of  wood  10ft.  7in.  long, 
4ft.  wide,  and  5ft.  lOin.  high  ?  Ans.  246||ft. 

15.  How  many  garments,  each  containing  4yd.  2qr.  3na., 
can  be  made  from  112yd.  2qr.  of  cloth? 

16.  Bought  lgal.  2qt.  lpt.  of  wine  for  $  1.82 ;  what  would 
be  the  price  of  a  hogshead  ?  Ans.  $  70.56. 

17.  Bought  125jyd.  of  lace  for  $  15.06  ;  what  was  the  price 
of  1  yard?  Ans.  $0.12. 

18.  What  cost  17cwt.  3qr.  of  wool,  at  $35.75  per  hundred- 
weight? Ans.  $634.5 62 J. 

19.  What  cost  7hhd.  47gal.  of  wine,  at  $87.25  per  hogs- 
head ?  Ans.  $  675.84F%. 

20.  How  many  solid  feet  in  a  stick  of  timber  34ft.  9in.  long, 
lft.  3in.  wide,  and  1ft.  6in.  deep  ?  Ans.  65.15625ft. 

21.  If  18yd.  lqr.  of  cloth  cost  $  36.50,  what  is  the  price  of  1 
yard  ?  Ans.  $  2.00. 


216  MISCELLANEOUS   EXAMPLES. 

22.  If  $  477.72  be  equally  divided  among  9  men,  what  will 
be  each  man's  share  ?  Ans.  $  53.08. 

23.  A  man  bought  a  barrel  of  flour  for  $  5.375,  7gal.  of  mo- 
lasses for  $  1.78,  9gal.  of  vinegar  for  $  1.1875,  lgal.  of  wine  for 
$  1.125,  141b.  of  sugar  for  $  1.275,  and  51b.  of  tea  for  $  2.625  ; 
what  did  the  whole  amount  to  ?  Ans.  $  13.367^. 

24.  A  man  purchased  3  loads  of  hay;  the  first  contained  2| 
tons,  the  second  3  J  tons,  and  the  third  lT*g-  tons ;  what  was  the 
value  of  the  whole,  at  $  17.625  a  ton  ?       Ans.  $  128.882{£ . 

25.  How  many  hogsheads  of  water  will  it  take  to  fill  a 
cistern  which  is  15.25  feet  long,  8.4  feet  wide,  and  10  feet 
deep  ?  Ans.  152hhd.  6T6Tgal. 

26.  At  $13,625  per  cwt.,  what  cost  3cwt.  2qr.  71b.  of 
sugar?  Ans.  $48,641^. 

27.  At  $  125.75  per  acre,  what  cost  37 A.  3R.  35rd.  ? 

Ans.  $  4774.570TV 

28.  At  $11.25  per  cwt.,  what  cost  17cwt.  2qr.  211b.  of 
rice?  Ans.  $199,237^. 

29.  What  cost  7  J  bales  of  cotton,  each  weighing  3.37cwt.,  at 
$  9.374-  per  cwt.  ? 

30.  What  cost  7hhd.  49gal.  of  wine,  at  $97,625  per  hogs- 
head ?  Ans.  $  759.305§f . 

31.  What  cost  7yd.  3qr.  3na.  of  cloth,  at  $4.75  per 
yard?  Ans.  $37.703£. 

32.  What  cost  27T.  15cwt.  lqr.  3£lb.  of  hemp,  at  $183.62 
per  ton  ?  ••  Ans.  $  5098.071£££. 

33.  What  is  the  cost  of  constructing  a  railroad  17m.  3fur. 
15rd.,  at  $  1725.875  per  mile  ?  Ans.  $  30067.978f  £, 

34.  When  $  624.53125  are  paid  for  17A.  3R.  15p.  of  land, 
what  is  the  cost  of  one  acre  ? 

35.  Paid  $494.53125  for  19T.  locwt.  2qr.  141b.  of  hay; 
what  was  the  cost  per  ton  ?  Ans.  $  24.9993y/T. 

36.  How  much  land,  at  $  40  per  acre,  can  be  obtained  for 
$  1004.75  ?  Ans.  25A.  OR.  19p. 

37.  How  many  cords  of  wood  can  be  put  into  a  space  20.5 
feet  long,  12.75  feet  wide,  and  7.6  feet  high  ? 

Ans.  15  cords  66£j  cubic  feet 

38.  How  many  bushels  of  corn  at  $  0.62  J  per  bushel  must  a 
farmer  exchange  for  31  yards  of  sheeting  at  $  0.08£  per  yard, 
and  7  J  yards  of  broadcloth  at  S  2.75  per  yard  ?     Ans.  37^. 


CIRCULATING   DECIMALS.  217 

39.  I  have  expended  $  42.875  for  a  quantity  of  grain,  -^  of 
it  being  corn,  at  $  0.75  a  bushel ;  ^  of  it  wheat,  at  $  2  a 
bushel ;  and  the  balance  oats,  at  $  0.40  a  bushel,  to  the  amount 
of  $  3.50.  Eequired  the  number  of  bushels  of  each  kind  pur- 
chased. , 

40.  If  a  mason,  in  constructing  a  drain  250.35  feet  long, 
begin  with  a  width  of  8  inches,  and  increase  T5^  of  an  inch 
in  every  foot  of  length,  how  many  times  the  width  of  the  be- 
ginning of  the  drain  will  its  end  be  ?  Ans.  16.646875. 

41.  A  gentleman  gave  \  of  his  property  to  his  son  James ; 
£  of  it  to  his  son  "William ;  \  of  the  remainder  to  his  daughter 
Mary ;  and  the  balance  to  his  wife.  It  appeared  that  Mary 
received  $  2243.26  less  than  James.  What  was  the  amount 
divided,  and  how  much  did  each  receive  ? 

Ans.  Amount,    $13459.56;    James,    $3364.89;    William, 
$4486.52;  Mary,  1121.63  ;  wife,  $  4486.52. 


CIRCULATING  DECIMALS. 

281.  A  Circulating  Decimal  is  a  decimal  in  which  one 
or  more  figures  are  continually  repeated  in  the  same  order. 
Thus,  in  reducing  ^  to  an  equivalent  decimal,  on  annexing 
ciphers  and  dividing  by  the  denominator,  the  result  obtained, 
.333-)-?  is  a  circulating  decimal ;  for,  however  far  the  division 
might  be  carried,  the  same  figure  would  continue  to  be  repeated 
without  the  decimal  terminating. 

Such  decimals  are  sometimes  called  infinite,  or  repeating ; 
and,  for  sake  of  distinguishing,  those  decimals  that  terminate 
are  sometimes  termed  finite. 

282.  A  repetend  is  a  figure,  or  a  series  of  figures,  contin- 
ually repeated.  To  mark  a  repetend,  a  point  (.)  is  placed  over 
a  single  repeating  figure,  or  over  the  first  and  last  of  a  series 
of  repeating  figures.  Thus,  in  .3,  the  point  denotes  that  the  3 
is  a  repetend;  and  in  .72,  that  the  72  is  a  repetend. 

283.  A  single  repetend  is  one  in  which  only  one  figure  is 

N    19 


218  CIRCULATING   DECIMALS. 

repeated;  as  in  .1111— |—  denoted  by  .1;  and  2222-f-,  denoted 
by  2. 

284.  A  compound  repetend  is  one  in  which  the  same  set 
of  figures  is  repeated;  as  in  .135135— 1—?  denoted  by  .135,  and 
.30363036+,  denoted  by  3086. 

285.  A  pure  repetend  is  one  which  contains  only  the  figures 
of  the  repetend ;  as,  .3,  .02,  and  .123. 

286.  A  mixed  repetend  is  one  in  which  a  repetend  is 
preceded  in  the  same  fraction  by  one  or  more  figures. 
The  figures  preceding  the  repetend  are  called  the  finite  part. 
Thus,  .416  is  a  mixed  repetend,  of  which  the  figure  6  is  the 
repetend,  and  the  figures  41  the  finite  part ;  also,  1.728  is  a 
mixed  repetend,  of  which  the  figures  28  are  the  repetend,  and 
the  figures  1.7  the  finite  part. 

287.  A  perfect  repetend  is  a  pure  repetend  containing  the 
same  number  of  figures  as  there  are  units  in  its  denominator 
less  one.  Thus,  ^  reduced  to  a  decimal  gives  .142857,  which, 
as  it  contains  as  many  figures  as  there  are  units  in  the  denomi- 
nator, 7,  less  one,  is  a  perfect  repetend. 

288.  Similar  repetends  are  those  which  begin  at  the  same 
distance  from  the  decimal  point;  as  .3  and  .6;  or  5.123  and 
3.478. 

289.  Dissimilar  repetends  are  those  which  begin  at  differ- 
ent distances  from  the  decimal  point ;  as  .986  and  .4625 ;  or 
.5920  and  .0423436. 

290.  Conterminous  repetends  are  those  which  terminate  at 
the  same  distance  from  the  decimal  point;  as  .631  and .465, or 
.0753  and  .4752. 

291.  Similar  and  conterminous  repetends  are  those  which 
both  begin  and  end  at  the  same  distance  from  the  decimal 
point ;  as  .354  and  .425  ;  or  .5757  and  5723. 

292.  Repetends  always  arise  from  common  fractions,  which, 
when  in  their  lowest  terms,  contain  in  their  denominator  other 
factors  than  2  and  5.  For  when  a  common  fraction  is  in  its 
lowest  terms,  its  numerator  and  denominator  are  prime  to  each 


CIRCULATING   DECIMALS.  219 

other  (Art.  219),  and  the  annexing  of  one  or  more  ciphers  to 
the  numerator  makes  the  same  a  multiple  of  10,  but  does  not 
render  it  divisible  by  any  factor,  except  2  and  5,  the  factors  of 
10.  Therefore,  when  the  denominator  of  a  fraction,  in  its 
lowest  terms,  contains  other  factors  than  those  of  10,  the  deci- 
mal resulting  from  dividing  the  numerator  with  ciphers  an- 
nexed, will  not  terminate,  but  will  contain  one  or  more  figures 
that  constantly  repeat. 

293.  A  pure  repetend  is  always  equivalent  to  a  common 
fraction  whose  numerator  is  the  repeating  figure  or  figures,  and 
whose  denominator  as  many  places  of  nines  as  there  are  repeat- 
ing figures.  For,  by  reducing  ^  to  a  decimal,  we  obtain  as  its 
equivalent  the  repetend  .1;  and  since  .1  is  equivalent  to  ^, 
.2  will  be  equivalent  to  f ,  .3  to  §,  and  so  on,  till  .9  is  equal  to  f 
or  1.  Again,  ^,  and  F^,  being  reduced,  give  .01,  and  001 ; 
that  is,  -fa  =  .01,  and  FJF  =  .001 ;  therefore,  ¥2F  =  .02,  and 
^f  3-  =  ^02,  anc*  so  on  5  tne  same  principle  holding  true  in  all 
like  cases. 

294.  A  mixed  repetend  is  equivalent  to  a  complex  decimal 
(Art.  278),  or  to  a  complex  fraction.  Thus,  the  mixed  repe- 
tend .2412  is  equivalent  to  the  mixed  decimal  .24^§,  which  is 

2412 
equal  to  the  complex  fraction  -^—p 

295  •  JRepetends  are  of  the  same  denomination  only  when 
they  are  similar  and  conterminous.  For  then  alone,  by  hav- 
ing a  common  denominator,  do  they  express  fractional  parts  of 
the  same  unit. 

REDUCTION  OF  REPETENDS. 

296.  To  reduce  a  repetend  to  an  equivalent  common  frac- 
tion. 

Ex.  1.  Reduce  .123  to  an  equivalent  common  fraction. 

Ans.  jfa 
.    .       operation.  We  write  the  figures  of  the  given 

.123  =  -J§§  =  ££g  Ans.     repetend  with    the    decimal    point 

omitted  for  the  numerator,  and  as 
many  nines  as  places  in  the  repetend  for  the  denominator,  of  a  com- 
mon fraction  (Art.  293),  and  obtain  Iff,  which,  reduced  to  its  low- 
est terms,  =  ^^,  the  answer  required. 


220  CIRCULATING  DECIMALS. 

2.  Reduce  .138  to  an  equivalent  common  fraction. 

Ans.  -sV 

operation.  The  mixed  repetend  .138 

\  3  8  is   equivalent   to  the  mixed 

.13  8  ==— -^  =  *§§■  =  &  Ans.     decimal    .13$     (Art.    294), 

1  0  U  which  we  readily  change  to 

the  form  of  a  complex  frac- 
tion by  erasing  the  decimal  point  and  writing  the  denominator,  100, 

1 3  8. 
which  is  understood ;    and  thus  obtain  ~g ,  which,  reduced  to  its 

simplest  form,  gives  -^ ,  the  answer  required.     Hence, 

If  the  given  repetend  be  simple,  make  the  repeating  figure  or  figures 
the  numerator,  and  take  as  many  nines  as  the  repetend  has  figures  for 
the  denominator. 

If  the  given  repetend  be  mixed,  change  it  to  an  equivalent  complex 
fraction,  and  that  fraction  to  its  simplest  fo)~m. 

Note.  —  Any  circulating  decimal  may  be  transformed  into  another  decimal, 
having  a  repetend  of  the  same  number  of  figures;  as,  .78  =  .787,  and  .534  = 
.5345  =  .53453.  Thus,  when  such  expressions  as  12.5  or  17.56  occur,  they  may 
be  also  transformed;  as  12.5  =  12.52,  and  17.56  =  17.567  =  17.5675,  &c. 

Examples. 

3.  Required  the  common  fraction  equal  to  .6.     Ans.  |  =  §. 

4.  Reduce  1.62  to  its  equivalent  mixed  number. 

Ans.  Iff. 

5.  Change  .53  to  an  equivalent  common  fraction. 

6.  What  common  fraction  is  equivalent  to  .769230  ? 

Ans. -if. 

7.  What  common  fraction  is  equivalent  to  .5925  ? 

8.  Change  31.62  to  an  equivalent  mixed  number. 

Ans.  31ff 

9.  Reduce  .008497133  to  an  equivalent  common  fraction. 

AnS.  -g-y^g-. 

297.  To  determine  the  kind  of  decimal  to  which  a  given 
common  fraction  can  be  reduced. 

Ex.  1.  Required  to  find  whether  the  decimal  equal  to  xtVo 
be  finite  or  circulating ;  and  if  finite,  of  how  many  places  the 
decimal  will  consist.  Ans.  Finite,  of  4  places. 


CIECULATING   DECIMALS.  221 

operation.  We      reduce 

3  the    given  frac- 

tWu  —  l\  =  9  v  o   v  o   v  9  ^  -1  8  7  5-     tion  to  its  lowest 

J  X  ^  X  &  X  ^  terms,  and  then 

resolve  the  de- 
nominator, 16,  of  the  fraction  obtained,  T8g,  into  its  prime  factors, 
which  we  find  to  be  2  X  2  X  2  X  2.  Now,  since  the  denominator 
contains  no  prime  factor  other  than  2  or  5,  it  is  evident  that,  by  an- 
nexing ciphers  to  the  numerator,  3,  and  dividing  by  the  denomina- 
tor, 16,  the  decimal  arising  will  terminate,  and  thus  be  finite  (Art. 
292). 

Since,  in  reducing  a  common  fraction  to  its  equivalent  decimal,  we 
annex  ciphers  to  the  numerator  and  divide  by  the  denominator  (Art. 
278),  every  10,  or  2  and  5,  that  enter  into  the  denominator  as  fac- 
tors must  produce  one  decimal  place,  and  no  more,  and  therefore 
every  other  factor  2  or  5  must  give  one,  and  only  one,  decimal  place. 
The  denominator,  16,  contains  only  the  factor  2  taken  4  times,  or  24 ; 
and  the  exponent  of  the  2  indicates  that  the  decimal  equivalent  to 
A  must  contain  exactly  4  decimal  places,  which  we  verify  by  reduc- 
ing the  ^q  to  its  equivalent  decimal,  .1875. 

2.  Find  whether  the  decimal  equal  to  ■&%%  be  finite  or  cir- 
culating ;  and  if  circulating,  of  how  many  places  the  finite  part, 
if  any,  and  the  circulating  part,  will  each  consist. 

Ans.  Circulating :  the  finite  part,  2  places  ;  the  repetend,  6 
places. 

We  reduce 


OPERATION. 


4  7  .  the       given 

iWir  =  ^  =  2v  5  v  ,  v7  =.13428571.   fraction     to 

^^°x'  its       lowest 

terms,  and 
obtain  $fc.  The  denominator,  350,  =  2  X  5  x  5X  7,  contains  a 
prime  factor,  7,  other  than  2  and  5 ;  therefore  the  decimal  equivalent 
to  -g^j.  will  contain  a  repetend ;  and  as,  of  the  factors  2  and  5,  the 
higher  exponent  of  either,  that  of  5,  is  2,  the  decimal  will  have  2 
finite  places  before  the  repetend  commences.  This  we  verify  by  re- 
ducing $fc  to  its  equivalent  decimal,  .13428571.  Hence,  to  deter- 
mine whether  the  decimal  to  which  a  given  common  fraction  can  be 
reduced  is  finite  or  circulating,  and  the  number  of  finite  decimal 
places,  if  any, 

Having  reduced  the  given  common  fraction  to  its  lowest  terms,  resolve 
the  denominator  into  its  prime  factors.  If  these  factors  be  not  other  than 
2  or  5,  the  decimal  icill  be  finite;  if  other  prime  factors  occur  with  2 
or  5,  the  decimal  will  be  a  mixed  repetend  ;  and  if  neither  2  nor  5 
occurs  as  factor,  the  decimal  will  be  a  pure  repetend. 

Whichever  factor  2  or  5  occurs  in  the  denominator  with  the  higher 
exponent  will  by  its  exponent  denote  the  number  of  finite  decimal 
places. 

19* 


222  CIRCULATING   DECIMALS. 

Note.  —  The  number  of  figures  of  which  a  repetend  will  consist  may  be  dis- 
covered by  dividing  1  with  ciphers  annexed  by  the  factors  other  than  2  or  5 
of  the  denominator,  until  there  is  a  remainder  1.  Thus,  if  it  be  required  to 
discover  the  number  of  figures  in  the  repeating  part  of  the  decimal  equivalent 
to  g4^j,  we  divide  1  with  ciphers  annexed  by  7,  the  only  prime  factor  in  the 
denominator  other  than  2  or  5,  until  there  is  a  Remainder  of  1,  which  occurs 
after  the  sixth  division,  thereby  indicating  that  the  repeating  part  will  consist 
of  six  figures.  We  have  seen  that  these  must  be  preceded  by  two  places  of 
finite  decimals,  so  that  the  mixed  repetend  equal  to  ^  must  consist  of  eight 
places  in  all. 

Examples. 

3.  To  what  kind  of  a  decimal  can  ^  be  reduced  ? 

Ans.  A  pure  repetend,  of  2  places. 

4.  How  many  places  of  decimals,  finite  and  repeating,  will  be 
required  to  express  T§| ^  ? 

Ans.  5  places  ;  3  finite  and  2  repeating. 

5.  To  what  kind  of  a  decimal  can  J-?  J  be  reduced? 

6.  Keduce  13J-J  to  a  mixed  repetend.  Ans.  13.37. 

7.  Change  T£f f^  to  a  mixed  repetend.    Ans.  .008407133. 

8.  Of  how  many  figures  will  the  repetend  consist  that  corre- 
sponds to  £§  ?  Ans.  28  figures. 

TRANSFORMATION  OF  REPETENDS. 

298.  Any  finite  decimal  may  be  considered  as  a  mixed 
repetend  by  making  ciphers  continually  recur ;  thus,  .42  = 
.420  =  .4200  =  .42000,  &c. 

299.  Any  circulating  decimal  may  be  transformed  into  an- 
other having  the  same  number  of  repeating  figures;  thus,  .127 
=  .1272  =  .12727,  &c. 

300 •  Any  circulating  decimal  having  as  repetend  any  num- 
ber of  figures  may  be  transformed  to  another  having  twice  or 
thrice  that  number  of  figures,  or  any  multiple  thereof;  thus, 
.5925,  having  a  repetend  of  three  figures,  may  be  transformed 
to  one  having  6, 9, 12,  &c  places ;  therefore  .5925  =  .5925925 
=  5925925925  =  5925925925925925,  &c 

301 .  The  value  of  a  decimal  is  not  changed  by  any  of  the 
above  transformations,  as  may  be  seen  by  reducing  the  given 


►  Ans. 


CIRCULATING   DECIMALS.  223 

repetends  to  their  equivalent  common  fractions  (Art.  296)  and 
comparing  them  together.  Hence,  they  can  be  used  in  making 
dissimilar  repetends  similar  and  conterminous. 

302.  To  make  any  number  of  dissimilar  repetends  similar 
and  conterminous. 

Ex.  1.  Make  similar  and  conterminous  9.167,  14.6,  3.165, 
12.432,  8.181,  and  1.307. 

OPERATION. 

Dissimilar.  Similar.  Similar  and  Conterminous. 

9.16  7  =      9.167  6  =      9.61767676] 

14.6  =14.600  =14.60000000 

3.1  6  5  ma      3.1  6  5  =      3.1  6  5  5  5  5  5  5 

1  2.4  32  =1  2.4  3243  =  1  2.4  3243243 

8.1  8  i  =      8.18  18  =      8.1  8  i  81818 

1.3  0  7  =      1.3  0  7  3  6  =      1.3  0  7  3  0  7  3  6  , 

We  make  the  finite  mixed  decimal,  14.6,  a  mixed  repetend  by 
annexing  recurring  ciphers,  and  make  it  and  all  the  given  repetends 
similar,  by  extending  the  figures  to  the  right,  so  that  the  circulating 
part  of  each  may  begin  at  the  same  distance  from  th^  decimal  point 
as  does  that  repetend  which  is  preceded  by  the  most  finite  decimal 
places.  Then,  to  make  conterminous  the  repetends  that  have  thus 
been  rendered  similar,  as  some  of  them  consist  of  1,  some  of  2,  and 
the  others  of  3  places,  we  extend  the  repeating  figures  of  each  repe- 
tend till  those  of  each  occupy  as  many  places  as  there  are  units  in  the 
least  common  multiple  of  1,  2,  and  3,  which  is  6.  Hence,  to  make 
dissimilar  repetends  similar  and  conterminous, 

Transform  the  given  repetends  so  that  the  circulating  parts  shall  com- 
mence at  the  same  distance  from  the  decimal  point,  and  shall  consist  of 
as  many  circulating  places  as  there  are  units  in  the  least  common  mul- 
tiple of  the  number  of  repeating  figures  found  in  the  given  decimals. 

Examples. 

2.  Make  3.671,  1.007i,  8.52,  and  7.616325  similar  and  con- 
terminous. 

3.  Make  1.52,  8.7156,  3.567,  and  1.378  similar  and  conter- 
minous. 

4.  Make  .0007,  .141414,  and  887. i  similar  and  conterminous. 

5.  Make  .3123,  3.27,  and  5.02  similar  and  conterminous. 

6.  Make  17.0884,  1563.0929,  and  15.i2345  similar  and  con- 
terminous. 


224  CIRCULATING   DECIMALS. 


ADDITION  OF   CIRCULATING  DECIMALS. 

303.     Ex.  1.    Add  2.765r  7.16674,  3.671,  .7,  and  .i728 
together.  Ans.  14.55436. 

operation.  Having  made  the  given 

Dissimilar.               Similar  and  Conterminous.  repetends  similar  and  COn- 

2.7  6  5          =      2.7  6  5  6  5  terminous  (Art.  302),  we 

716674  =     716674  a<^    ^    m    aa*dition    of 


3.6  7  1         =     3.67  13  6 


whole  numbers,  and  obtain 
14.55433.    The  right-hand 


.7  =        .77777  figure  of  this  result  we  in- 

-J728  17281  crease  by  such  a  number 

- — . as  would  have  been  car- 

Ans.     1  4.5  5  4  3  6  ried,  if  the  repetends  had 

been  continued  farther  to 
the  right.  In  that  case  we  should  have  had  to  carry  3  after  finding 
the  amount  of  the  first  left-hand  column  of  the  repetends  continued. 
We  therefore  increase  the  sum  as  first  found,  and  thus  have  the  true 
amount  as  in  the  operation,  14.55436. 

Rule.  —  Make  the  given  repetends,  wlu  n  dissimilar,  similar  and  con- 
terminous. A(M  as  in  addition  of  finite  decimals,  observing  to  increase 
the  repetend  of  the  amount  by  the  number,  if  any,  to  be  carried  from  the 
left-hand  column  of  the  repetends. 

Examples. 

2.  Add  3.5,  7.65i,  1.765,  6.173,  51.7,  3.7,  27.63i,and  1.003 
together.  Ans.  103.2591227. 

3.  Reduce  -J,  -f,  and  £  to  decimals,  and  find  their  sum. 

4.  Find  the  sum  of  27.56,  5.632,  6.7,  16.356,  .11,  and 
6J234.  Ans.  63.1690670868888. 

5.  Add  together  .165002,  31.64,  1.06,  .34634,  and  13. 

6.  Add  together  .87,  .8,  and  .876.  Ans.  2.644553. 

7.  Required  the  value  of  .3  +  .45  -|-  .45  -f-  .35 i  +  .6468 
+  .6468  -f  .6468,  and  .6468.  Ans.  4.1766345618. 

8.  Find  the  value  of  1.25  +  3.4  +  .637  +  7.885  -f  7.875 
+  7.875  +  11.X.  Ans.  40.079360724. 

9.  Add  together  131.613,  15.001,  67.134,  and  1000.63. 
10.  Find  the  value  of  5.i6345  +  8.6381  +  3.75. 

Ans.  17.55919120847374090302. 


(JIKCULATING   DECIMALS.  225 


SUBTRACTION  OF  CIRCULATING  DECIMALS. 

304.  Ex.  1.     From  87.1645  take  19.479167. 

Ans.  67.685377. 
operation.  Having  made  the  repe- 

Dissimilar.^  Similar  and  Conterminous.      ten(js  s;imlar  an(J  conter- 

8  7.1  6  4  5  =8  7.1  6  4  5  4  5         minous,  we  subtract  as  in 

H479H7   -      iQ4.7qifi7         whole  numbers,  regarding, 
1  y.4  I  J  1  b  (    —      1  a.4=  /  J  1  b  /  however,   the    right-hand 

Ans.     6  7.6  8  5  3  7  7         %ure  of  the  subtranend 

as  increased  by  1,  since  1 

would  have  been  carried 

to  it  in  subtracting,  if  the  repetends  had  been  continued  farther  to 

the  right,  as  is  evident  from  the  circulating  part  of  the  subtrahend 

being  greater  than  that  of  the  minuend. 

Rule.  —  Make  the  repetends,  when  dissimilar,  similar  and  conter- 
minous. Subtract  as  in  subtraction  of  finite  decimals;  observing  to 
regard  the  repetend  of  the  subtrahend  as  increased  by  1,  when  it  exceeds 
that  of  the  minuend. 

Examples. 

2.  From  7.i  take  5.02.  Ans.  2.08. 

3.  From  315.87  take  78.0378.     Ans.  237.838072095497. 

4.  Subtract  |  from  f .  Ans.  .079365. 

5.  From  16.1347  take  11.0884.  Ans.  5.0462. 

6.  From  18.1678  take  3.27.  Ans.  14.8951. 

7.  From  3.123  take  0.7 i.  Ans.  2.405951. 

8.  From  f  take  T2T.  Ans.  .246753. 

9.  From  §  take  f .  Ans.  .i58730. 

10.  From  fr  take  fr.  Ans.  .i76470588235294i. 

11.  From  5.i2345  take  2.3523456. 

Ans.  2.7711055821666927777988888599994. 

MULTIPLICATION  OF  CIRCULATING  DECIMALS. 

305,  Ex.  1.  Multiply  .36  by  25.  Ans.  0929. 

operation.  "We    change    the 

'6h  ~  99  =  U;    lb  ~  F6  ~~  90"  their  equivalent  com- 

.     .  mon    fractions,   and, 

tt  Xff=  t&2<j  —  ith  =  -0929  Ans.    multiplying,     obtain 

A2o   -  T4A,   which, 
reduced  to  its  equivalent  decimal,  gives  .0925,  the  answer  required. 


226  CIRCULATING   DECIMALS. 

Rule.  —  Change  the  given  numbers  to  their  equivalent  common  frac- 
tions. Multiply  them  together,  and  reduce  the  product  to  its  equivalent 
decimal. 

Examples. 

2.  Multiply  87.32586  by  4.37.  Ans.  381.6140338. 

3.  Multiply  582.347  by  .03. 

4.  Multiply  3.145  by  4.297.  Ans.  13.5169533. 

5.  What  is  the  value  of  .285714  of  a  guinea?       Ans.  8s. 

6.  What  is  the  value  of  .461607i42857  of  a  ton  ? 

Ans.  9cwt.  Oqr.  23+lb. 

7.  What  is  the  value  of  .284931506  of  a  year  ? 

Ans.  104d. 


DIVISION  OF  CIRCULATING  DECIMALS. 

306.     Ex.  1.  Divide  .54  by  .15.  Ans.  3.506493. 

operation.  We    change 

54         6         •         1£        14         7  the  given  num- 


M  ~  99  ~  U;  -15  ~~  10  _  90  ~  45"         be™    J0  /heir 

equivalent  com- 

T6T  -fc  &  —  W  =  3?f  =  3.506493  Ans.    mon     fractions, 

and,      dividing, 

obtain  3^,  which,  reduced  to  its  equivalent  decimal,  gives  3.506493, 

the  answer  required. 

Rule.  —  Change  the  given  numbers  to  their  equivalent  common  frac- 
tions.    Divide,  and  reduce  the  quotient  to  its  equivalent  decimal. 

Examples. 

2.  Divide  345.8  by  .6.  Ans.  518.83. 

3.  Divide  234.6  by  .7. 

4.  Divide  13.51 69533  by  3.145.  Ans.  4.297. 

5.  Divide  381.6140338  by  4.37.  Ans.  87.32586. 

6.  Divide  .42857±  by  .625. 

7.  Find  the  value  of  2.370  -f-  4.923076.  Ans.  .48i. 

8.  Find  the  value  of  .09  -r-  .230769.  Ans.  .39. 

9.  Find  the  value  of  316.31015  -r-  .3. 

10.  Find  the  value  of  100006  -r-  .6. 

11.  Divide  .36  by  .25.     Ans.  1.422924901185770750988L 


CONTINUED   FRACTIONS.  227 


CONTINUED  FRACTIONS. 

307.     A  continued  fraction  is  a  fraction  having  for  its 
numerator  1,  and  for  its  denominator  a  whole  number  plus  a 
fraction  whose  numerator  is  1,  and  whose  denominator  is  a 
whole  number  plus  a  fraction,  and  so  on.     Thus, 
1 


3  +  1 


2  +  1 


5  +  1 


4  +  J-,  is  a  continued  fraction. 
The  partial  fractions  composing  the  parts  of  a  continued 
fraction  are  called  its  terms.    Thus,  in  the  fraction  given  above, 
h  h  h  ^c-  are  its  terms. 

308.  Continued  fractions  are  used  in  obtaining,  in  smaller 
numbers,  the  approximate  values  of  fractions  whose  terms, 
when  reduced  to  their  simplest  forms,  are  expressed  in  num- 
bers inconveniently  large. 

309.  To  transform  a  common  fraction  into  a  continued  frac- 
tion, and  to  find,  in  smaller  numbers,  its  approximate  values. 

Ex.  1.  Transform  -if  into  a  continued  fraction,  and  find  its 
several  approximate  values. 


OPERATION. 


19)60(3  TT  19         1 

57 v  Hence, 


60        3  +  1 


Ans. 


3)19(6  6  +  ^ 

18  1  I    • 

~1)3(3      ~3  = -g-,  1st  approx.  value, 

3  lx6  6    OA  i 

0"        (8X6)  +  1  =  19'  2d  appr0X'  Value' 

(6X3)  +  1       19  .  . 

(19X3)  +  3  =  60' the  °ngmal  Valu6' 

Dividing  both  terms  of  J$  by  the  numerator,  which  operation  -will 

not  change  the  value  expressed  (Art.  217),  the  fraction  becomes 

— ;  the  denominator  of  which  being  between  3  and  4,  the  value  of 


228  CONTINUED   FRACTIONS. 

the  given  fraction  must  be  between  J  and  l ;  and  neglecting  the  frac- 
tion -j8T,  for  the  present,  in  the  denominator,  we  have  ^  for  the  Jirst 
approximate  value.  This  approximation,  however,  is  greater  than 
the  true  value,  since  the  denominator,  3,  is  less  than  the  true  denom- 
inator 3T3¥.  We  therefore  divide  both  terms  of  the  fy  by  its  nu- 
merator, and  it  becomes  — -,  which  is  between  J  and  \.  By  neglecting 
the  fraction  ^  in  the  denominator,  and  taking  the  J  instead  of  the 
-fy,  we  have  - — : — z  =  jr  =  T6^  for  the  second  approximate  value  of 

the  given  fraction;  which  approximation  is  too  small,  since  in  the 
denominator,  instead  of  -j\,  we  used  J-,  which  is  greater  than  the  T87. 
If  we  now  include  in  the  calculation  the  remaining  partial  fraction 
J,  we  have  — = =  £$,  the  original  fraction. 

By  the  processes  of  the  operation  it  will  be  seen  that  the  Jirst  ap- 
proximate value  sought  was  obtained  by  disregarding  all  the  partial 
fractions  after  the  first,  the  second  approximate  value  by  disregard- 
ing all  the  partial  fractions  after  the  second,  &c. 

Rule.  —  Divide  the  greater  term  of  the  given  fraction  by  the  less, 
and  the  divisor  by  the  remainder,  and  so  on,  as  in  finding  the  greatest 
common  divisor.  The  quoth  nit  thus  found  will  be  the  denominators  of 
the  several  tei*ms  of  the  continued  fraction,  and  the  numerator  of  each 
will  be  1. 

For  the  first  approximation,  take  the  first  terms  of  the  continued 
fraction. 

For  the  second  approximation,  multiply  the  terms  of  the  first  ap- 
proximate fraction  by  the  denominator  of  the  second  term  of  the  con- 
tinued fraction,  adding  1  to  the  product  of  the  denominators. 

For  each  succeeding  approximation,  multiply  the  terms  of  the  ap- 
proximation last  found  by  the  denominator  of  the  next  term  of  the 
continued  fraction,  and  add  the  corresponding  terms  of  the  preceding 
approximation . 

Note  1.  —  When  the  fraction  given  is  improper,  the  true  approximations  ivill 
be  the  reciprocals  of  the  fractions  found  by  the  rule. 

Note  2.  —  In  a  series  of  approximations  the  first  is  smaller,  the  second 
larger,  and  so  on,  every  odd  fraction  being  smaller,  and  every  even  one  larger, 
than  the  given  fraction.  Each  successive  approximate  fraction,  however,  ap- 
proaches more  nearly  than  the  one  preceding  it  to  the  value  of  the  given 
fraction.  When  the  continued  fraction  indicates  many  approximations,  it  is 
generally  sufficient  for  ordinary  purposes  to  find  only  from  three  to  six  of 
them. 

Note  3.  —  A  continued  fraction  may,  for  convenience,  be  expressed  by  writ- 
ing its  terms  one  directly  after  another,  with  the  sign  plus  (+)  between  the 
denominators ;  thus,  the  continued  fraction  equivalent  to  ig  mav  be  written 

i+i+h- 


RATIO.  229 

Examples. 


2.  Transform  i%  into  a  continued  fraction. 

Ans. 


2  +  1 


8  +  i. 


3.  Transform  -2ey-  into  a  continued  fraction. 

4.  Find  the  approximate  values  of  f  f- .       Ans.  |-,  J,  §,  §,  f-f . 

5.  Find  the  first  five  approximate  values  of  Iffy- 

6.  Find  the  first  three  approximate  values  of  ff-f . 

Ans.  } ,  &  J,  or  2,  2J,  2£. 

7.  Find  the  first  six  approximate  values  of  . 

Ans.  tV>  A>  a^n  fo  -N&J&r- 

8.  What  are  the  first  four  approximate  values  of  1.27  ? 

Ans.  i,  |,  |,  tf>  M>  or  1,  1*,  lfc  1T3T,  1&. 


RATIO. 

31(h  Ratio  is  the  relation,  in  respect  to  magnitude  or 
value,  which  one  quantity  or  number  bears  to  another  of  the 
same  kind. 

31 1#  The  comparison  by  ratio  is  made  by  considering  how 
often  one  number  contains,  or  is  contained  in,  another.  Thus, 
the  ratio  of  10  to  5  is  expressed  by  2,  the  quotient  arising 
from  the  division  of  the  first  number  by  the  second,  or  it  may 
be  expressed  by  -fij  =  £,  the  quotient  arising  from  the  division 
of  the  second  number  by  the  first,  as  the  second  or  the  first 
number  shall  be  regarded  as  the  unit  or  standard  of  comparison. 
In  general,  of  the  two  methods,  the  first  is  regarded  as  the 
more  simple  and  philosophical,  and  therefore  has  the  preference 
in  this  work. 

Note.  —  Which  of  the  two  methods  is  to  be  preferred,  is  not  a  question  of 
so  much  importance  as  has  been  by  some  supposed,  since  the  connection  in 
which  ratio  is  used  is  usually  such  as  to  readily  determine  its  interpretation. 

312t     The  two  numbers  necessary  to  form  a  ratio  are  called 
N     20 


% 


230  RATIO. 

the  terms  of  the  ratio.  The  first  term  is  called  the  antecedent, 
and  the  last,  the  consequent.  The  two  terms  taken  together  are 
called  a  couplet ;  and  the  quotient  of  the  two  terms,  the  index 
or  exponent  of  the  ratio. 

313.  The  ratio  of  one  number  to  another  may  be  expressed 
either  by  two  dots  ( : )  between  the  terms  ;  or  in  the  form  of  a 
fraction,  by  making  the  antecedent  the  numerator  and  the  con- 
sequent the  denominator.  Thus,  the  ratio  6  miles  to  2  miles 
may  be  expressed  as  6  :  2,  or  as  § . 

314*  The  terms  of  a  ratio  must  be  of  the  same  kind,  or  such 
as  may  be  reduced  to  the  same  denomination.  Thus,  cents 
have  a  ratio  to  cents,  and  cents  to  dollars,  &c. ;  but  cents  have 
not  a  ratio  to  yards,  nor  yards  to  gallons. 

315.  A  simple  ratio  is  that  of  two  whole  numbers ;  as, 
3  :  4,  8  :  16,  9  :  36,  &c. 

316.  A  complex  ratio  is  that  of  two  numbers,  of  which  one 
or  both  are  fractional ;  as,  6  :  4£,  f  :  -^,  4  J  :  2£,  &c. 

317*  A  compound  ratio  is  the  product  of  two  or  more  ra- 
tios. Thus,  the  ratio  compounded  of  4  :  2  and  6  :  3  is  J  X  f 
=  2£  =  4,  or  4  X  6:2  X  3  =  4. 

A  compound  ratio  is  generally  expressed  by  writing  the  ra- 
tios composing  it,  in  a  column,  with  the  antecedents  in  one 

vertical  line,  and  the  consequents  in  another ;  thus,  fi  \  o  [•  ex- 
presses a  compound  ratio. 

Note.  —  If  a  ratio  be  compounded  of  two  equal  ratios,  it  is  called  a  dupli- 
cate ratio;  of  three  ratios,  a  triplicate  ratio,  &c. 

318*  A  ratio  is  either  direct  or  inverse,  A  direct  ratio  is 
the  quotient  of  the  antecedent  by  the  consequent ;  an  inverse 
ratio,  or  reciprocal  ratio,  as  it  is  sometimes  called,  is  the  quo- 
tient of  the  consequent  by  the  antecedent,  or  the  reciprocal  of 
the  direct  ratio.  Thus  the  direct  ratio  of  6  to  2  is  §■  or  3 ;  and 
the  inverse  or  reciprocal  ratio  of  6  to  2  is  f  or  ^-,  which  is  the 
same  as  the  reciprocal  of  3,  the  direct  ratio  of  6  to  2. 

Note  1.  —  One  quantity  is  said  to  vary  directly  as  another,  when  both  in- 
crease or  decrease  together  in  the  same  ratio ;  one  quantity  is  said  to  vary  in- 


RATIO.  231 

versely  as  another,  when  the  one  increases  in  the  same  ratio  as  the  other  de- 
creases. 

Note  2.  —  The  word  ratio,  when  used  alone,  means  the  direct  ratio. 

319.  When  the  antecedent  and  consequent  of  a  ratio  are 
equal,  the  ratio  equals  1,  and  is  called  that  of  equality.  Thus, 
the  ratio  of  6  :  6  =  f  =  1,  and  the  ratio  of  6  X  4  :  8  X  3 
=  J4  ===  1,  are  ratios  of  equality.  But  if  the  antecedent  is 
larger  than  the  consequent,  the  ratio  is  that  of  greater  inequal- 
ity, and  if  the  antecedent  is  smaller  than  the  consequent,  the  ratio 
is  that  of  less  inequality.  Thus,  the  ratio  of  15  :  5  =  \b-  =  3, 
is  a  ratio  of  greater  inequality ;  and  the  ratio  of  7  :  14  =  T7^ 
=  £,  is  a  ratio  of  less  inequality. 

320.  The  ratio  of  two  fractions  having  a  common  numerator 
is  the  same  as  the  inverse  ratio  of  their  denominators.  Thus, 
the  ratio  of  f-  :  §  is  f-  -r-  ■§  =  2,  which  is  the  inverse  ratio  of 
the  denominator  4  to  the  denominator  8. 

321  •  The  ratio  of  two  fractions  having  a  common  denomi- 
nator is  the  same  as  the  ratio  of  their  numerators.  Thus,  the 
ratio  of  f  :  f  is  f-  ~  f-  =  2,  which  is  the  ratio  of  the  numera- 
tor 6  to  the  numerator  3. 

322.  The  inverse  or  reciprocal  ratio  of  two  numbers  de- 
notes what  part  or  multiple  the  antecedent  or  dividend  is  of  the 
consequent  or  divisor.  Thus,  inquiring  what  part  of  4  is  3,  or 
what  part  3  is  of  4,  is  the  same  as  inquiring  the  inverse  or  re- 
ciprocal ratio  of  4  :  3.  The  inverse  ratio  of  4  :  3  is  f ,  and  3 
is  f  of  4. 

323.  In  order  to  compare  one  number  with  another,  by 
ratio,  it  is  necessary  that  they  should  not  only  be  of  the  same 
kind,  but  of  the  same  denomination.  Thus,  to  compare  2  days 
with  12  hours,  it  is  necessary  that  the  days  be  reduced  to  hours, 
before  we  can  indicate  the  ratio,  which  is  48  hours  :  12  hours. 

324.  If  the  antecedent  of  a  ratio  he  multiplied,  or  the  conse- 
quent divided,  the  ratio  is  multiplied.  Thus,  the  ratio  of  6  :  3 
is  2,  but  6  X  2  :  3  is  4 ;  or  6  :  3  —  2  is  4. 

325.  If  the  antecedent  of  a  ratio  be  divided,  or  the  conse- 
quent multiplied,  the  ratio  is  divided.  Thus,  the  ratio  of  18  :  6 
is  4,  but  18  ~  3  :  6  is  1 ;  or  18  :  6  X  3  =  1. 


232  RATIO. 

326.  If  both  the  antecedent  and  consequent  of  a  ratio  be 
multiplied  or  divided  by  the  same  number,  the  ratio  is  not 
altered.  Thus,  the  ratio  of  8  :  2  is  4 ;  of  8  X  2  :  2  X  2  is  4; 
and  of  8  -7-  2  :  2  -r-  2  is  4. 

EEDUCTION  AND  COMPARISON  OF  RATIOS. 

327.  Ratios,  being  of  the  nature  of  fractions,  may  be  reduced, 
compared,  and  otherwise  operated  upon  like  them. 

328.  To  reduce  a  ratio  to  its  lowest  terms. 

Ex.  1.     Reduce  18  :  9  to  its  lowest  terms.         Ans.  2:1. 

operation.  "VYe  cancel  in  the  two  terms  the 

18  :  9  =  JF8-  =  f  =  2  :  1.     common  factor  9,  and   obtain  \  = 

2:1,  the  answer.     Hence 

Cancel  in  the  given  ratio  all  factors  common  to  its  terms. 

Examples. 

2.  Reduce  to  its  lowest  terms  63  :  72.  Ans.  £. 

3.  Reduce  to  its  lowest  terms  66  :  24. 

4.  Reduce  to  its  lowest  terms  4X6X3:8X9X2. 

Ans.  £. 

5.  What  are  the  lowest  terms  of  19  X  5  X  2  X  3  :  15  X 
12  X  38? 

329.  To  reduce  a  complex  or  a  compound  ratio  to  a 
simple  one. 

Ex.  1.     Reduce  5£  :  £■  to  a  simple  ratio.  Ans.  22  :  3. 

operation.  "VVe     express     the 

5A  :*  =  #  =  *>=*?=  22  :SAns.     fiven    ra,tio    in    the 

2       *  3  6  TJ  form        0f       a       com_ 

plex  fraction,  which, 
changed  to  a  simple  fraction  (Art.  242),  and  reduced  to  its  lowest 
terms,  gives  %£■  =  22  :  3,  the  answer  required. 


8  *    5  ) 
2.  Reduce  n  \  0yl  >  to  a  simple  ratio.  Ans.  7  :  15, 


7:24^ 

OPERATION. 


3 

We  express  the  given  ratio  in  the  form  of  a  compound  fraction, 


RATIO.  233 


which,  reduced  to  a  simple  one  (Art.  329),  gives  T\  =  7  :  15,  the 
answer  required.  Hence,  to  reduce  a  complex  or  a  compound  ratio 
to  a  simple  one, 

Proceed  as  in  like  operations  with  fractions. 

Examples. 

3.  Eeduce  f  :  f  to  a  simple  ratio.  Ans.  35  :  24. 

4.  Eeduce  13^- :  27  to  a  simple  ratio.  Ans.  1  :  2. 

5.  Eeduce  6.25  :  3.125  to  a  simple  ratio.  Ans.  2  :  1. 

6.  Eeduce  ok  !  in  i  to  a  smiP*e  rati°«  Ans.  5  :  8. 

3:    6] 

7.  Eeduce      9  :  27  >to  a  simple  ratio.  Ans.  3  :  2. 

108  :  12  ) 


8.  Eeduce  7625  *.  25  5  i  to  a  simPle  ratio# 


Ans.  6  :  1. 


330.     To  find  the  ratio  of  one  number  to  another. 

Ex.  1.     Eequired  the  direct  ratio  of  108  to  9.      Ans.  12. 

operation.  Since  9  is  the  unit  or  standard  of 

108:9=-1g-^  =  12  Ans.       comparison,  we  make  it  the  conse- 
quent (Art,  111)  and  the  108  the 
antecedent  of  the  ratio,  and  obtain  i$&  =12  Ans. 

2.  Eequired  the  inverse  ratio  of  72  to  8.  Ans.  £. 

operation.  We  divide*  the  consequent 

72  :  8  inverted  =  ^  ==  i  Ans.         8  Dv  tne  antecedent  72,  or, 

which  is  the  same  thing,  find 
the  reciprocal  of  the  direct  ratio  of  72  :  8  (Art.  318),  by  inverting  its 
terms,  and  thus  obtain  -f%  =  ^  Ans.     Hence, 

The  direct  ratio  is  found  by  dividing  the  antecedent  by  the  consequent, 
and  the  inverse  ratio  by  dividing  the  consequent  by  the  antecedent. 

Note  1.  —  Eatios  expressed  by  fractions  having  different  denominators 
must  be  reduced  to  a  common  denominator,  in  order  to  be  compared;  and 
then  they  are  to  each  other  as  their  numerators  (Art.  323). 

Note  2.  —  When  a  ratio  is  expressed  in  terms  inconveniently  large  and 
prime  to  each  other,  we  may  find  the  approximate  values  of  the  ratio  ex- 
pressed in  smaller  numbers,  as  in  other  fractional  expressions  (Art.  309). 

Examples. 

3.  What  is  the  ratio  of  39  to  13  ?  Ans.  3. 

4.  What  is  the  ratio  of  2  yards  2  quarters  to  9  yards  ? 

2(1* 


234  RATIO. 

5.  What  is  the  ratio  of  21  gallons  to  J  of  a  hogshead? 

Ans.  1. 

6.  What  is  the  ratio  of  £  of  £  of  $  2  :  \  of  $  0.50  ? 

Ans.  f . 

7.  What  is  the  inverse  ratio  of  24  :  6  ?  Ans.  £. 

8.  What  part  of  36  is  4  ?  Ans.  £. 

9.  What  part  of  a  farm  of  94A.  2R.  16rd.  is  11  A.  3R.  ? 

10.  Which  is  the  greater,  the  ratio  of  17  to  9,  or  of  39  to 
19?  Ans.  39:19. 

11.  By  how  much  does  the  ratio  of  36  X  4  X  3  :  12  X  16 
X  2  exceed  that  of  60  -*-  (3  X  5)  :  20  X  2  -r-  8  ?     Ans.  £g. 

12.  What  is  the  inverse  ratio  of  .02  :  2.503  ? 

13.  Which  is  the  greater,  the  ratio  of  \  of  \  :  -J  of  £,  or  that 
of5:4? 

14.  The  height  of  Bunker  Hill  Monument  is  220  feet,  and 
that  of  the  great  pyramid,  Egypt,  500  feet ;  what  is  the  ratio 
of  the  height  of  the  former  to  that  of  the  latter  ?       Ans.  ££. 

15.  A  certain  farm  contains  180  acres,  and  the  township  of 
which  it  forms  a  part  is  36  square  miles  in  extent.  What  is 
the  ratio  of  the  latter  to  the  former  ? 

16.  Find  approximate  values  for  the  ratio  of  4900  to  11283. 

Ans.  i,  f ,  ft  f  f>  tW,  &c. 

17.  The  ratio  of  the  circumference  of  a  circle  to  its  diameter 
is  3.141592.     Required  approximate  values  for  this  ratio. 

Ans.  3,  -V-,  m>  Hh  &c.,  or  3,  3|,  3TV\,  3T\%  &c. 

ANALYSIS  BY  EATIO. 

331.  Operations  by  analysis  may  often  be  much  abridged 
by  ratio.  Thus,  frequently,  it  is  more  convenient  to  multiply 
or  divide  by  the  ratio  a  number  bears  to  a  unit  of  the  same 
kind,  than  to  multiply  or  divide  by  the  number  itself. 

This  form  of  analysis  is  much  used  by  business  men  ;  and, 
like  that  by  aliquot  parts  (Art.  114),  is  sometimes  called  Prac- 
tice. 

Examples. 

1.  What  cost  14  tons  15cwt.  3qr.  201b.  of  iron,  at  $  60  a 
ton?  Ans.  $887.85. 


KATIO.  235 

OPERATION. 

$  60.00  =  cost  of  1  ton. 
14 


$  840.00  =t  "  14  tons. 

(lOcwt. :  1  ton  =  £)  ;  £  of  $  60  =  30.00  =s  «  lOcwt. 

(5cwt. :  lOcwt.  =  £) ;  £  of  $  30  =  15.00  *  "  5cwt. 

(2qr.  :  5cwt.  et  TV) ;  TV  of  $  15  =  1.50  =  "  2qr. 

(lqr.  :  2cwt.  =  £) ;  £  of  $  1.50  =  0.75  =*  «  lqr. 

(151b.  :  3qr.  ==  |) ;  $  of  $  2.25  ==  0.45  =  «  151b. 

(51b.  :  151b.  =  i)  ;  £  of  $  0.45  =  0.15  s  «  51b. 

Ans.     $887.85=     "     14T.  3qr.  201b. 

Since  1  ton  costs  $  60,  14  tons  will  cost  14  times  $  60,  or  $  840. 
15c\vt.  =  lOcwt.  -f-  5cwt.  Since  the  ratio  of  lOcwt.  to  1  ton  or 
20cwt.  =  -|,  lOcwt.  will  cost  ^  as  much  as  1  ton,  or  $  30  ;  and  as  the 
ratio  of  5cwt.  to  lOcwt.  =  J,  5cwt.  will  cost  £  as  much  as  lOcwt.  or 
$  15.  3qr.  =  2qr.  -\-  lqr.  Since  the  ratio  of  2qr.  to  5cwt.  or 
20qr.  =  ^q,  2qr.  will  cost  X  as  much  as  5c wt.,  or  $  1.50 ;  and  as 
the  ratio  of  lqr.  to  2qr.  =  ^,  lqr.  will  cost  i  as  much  as  2qr.,  or 
$0.75.  20lb.  =  15lb.  +  5lb.  Since  the  ratio  of  15lb.  to  3qr.  or 
75lb.  =  ^,  15lb.  will  cost  -^  as  much  as  3qr.,  or  $0.45;  and  as  the 
ratio  of  5lb.  to  15lb.  =  £,  5lb.  will  cost  £  as  much  as  15lb.,  or 
$  0.15.  The  cost  of  the  several  parts  equals  the  cost  of  the  whole,  or 
$887.85,  Ans. 

2.  What  is  the  value  of  17  acres  3  roods  35  rods  of  land, 
at  $  80  per  acre  ?  Ans.  $  1437.50. 

3.  What  cost  16cwt.  3qr.  101b.  of  guano,  at  $  2.50  per  cwt.  ? 

4.  What  cost  27cwt.  lqr.  201b.  of  coffee,  at  $14  per 
cwt.  ?  Ans.  $  384.30. 

5.  If  1  yard  of  cloth  cost  $5.60,  what  will  7yd.  3qr.  2na. 
cost?  Ans.  $44.10. 

6.  What  cost  7  tons  13cwt.  2qr.  of  hay,  at  $  20  per  ton  ? 

7.  What  cost  99bu.  lpk.  4qt.  of  wheat,  at  $1.92  per 
bushel?  Ans.  $191.80. 


OPERATION. 

$1.92     = 
100 

cost  of  lbu. 

(2pk. 
(4qt: 

:lbu. 
:2pk. 

=  1); 

$  192.00     = 
J  of  $1.92  =  0.96  >  = 
£of  $0.96  =  0.24)  = 

u 

100bu. 

2pk. 

4qt. 

Ans.     $191.80     = 

a 

99bu.  lpk.  4qt 

236  RATIO. 

The  quantity  being  nearly  100  bushels,  we  find  the  cost  of  100 
bushels  by  annexing  two  ciphers  to  $  1.92,  the  cost  of  1  bushel,  and 
obtain  $192,  from  which  we  subtract  the  cost  of  2pk.  4qt.,  the  differ- 
ence of  quantity  between  that  given  and  100  bushels;  the  cost  of 
2pk.  =  $0.96;  and  that  of  4qt.  =  $0.24;  $192  —  $0.96  -j- 
$0.24  =  $191.80  Ans. 

8.  What  cost  19yd.  3qr.  2na.  of  cloth,  at  $  4.40  per  yard  ? 

Ans.  $87.45. 

9.  How  much  must  be  paid  for  24A.  3R.  20p.  of  land,  at 
$  32  per  acre  ? 

10.  How  much  must  be  paid  for  1991b.  12oz.  of  butter,  at 
$  0.30  per  lb.  ?  Ans.  $  59.925. 

11.  What  cost  714  yards  of  broadcloth,  at  15s.  6d.  per 
yard  ?  Ans.  553£.  7s. 

12.  How  much  must  be  paid  for  the  services  of  a  man  2y. 
9mo.  15da-,  at  $  450  yer  year  ?  Ans.  $  1256.25. 

13.  If  1  acre  of  land  cost  $  80.50,  what  will  25  acres  2  roods 
35  rods  cost  ?  Ans.  $  2070.35-f. 

14.  What  cost  4981b.  of  tea,  at  2s.  6d.  per  lb.  ? 

15.  If  lcwt.  2qr.  121b.  of  alum  can  be  purchased  for 
$  4.05,  how  much  can  be  purchased  for  $  28.35  ? 

Ans.  lOcwt.  3qr.  91b. 

OPERATION. 

$28.35:  $4.05  =  7; 
lcwt.  2qr.  121b.  X?  =  lOcwt.  3qr.  91b.  Ans. 

Since  the  ratio  of  $  28.35  to  $  4.05  =  7,  $  28.35  will  purchase  7 
times  as  much  as  $4.05.  By  multiplying  what  the  latter  will  pur- 
chase by  the  ratio,  we  have  the  answer  required. 

16.  If  llgal.  3qt.  lpt.  of  molasses  cost  $5.83f,  what  will 
35gal.  2qt.  lpt.  cost?  Ans.  $17.51|-. 

17.  If  24yd.  3qr.  of  cloth  cost  $  49.50,  what  will  12yd.  lqr. 
2na.  cost? 

18.  If  17bu.  2pk.  4qt.  of  oats  be  paid  for  14bu.  3pk.  of  salt, 
what  quantity  of  oats  must  be  paid  for  73bu.  3pk.  of  salt  ? 

Ans.  88bu.  Opk.  4qt. 

19.  If  $  9.75  will  purchase  IT.  2cwt.  2qr.  151b.  of  coal,  how 
much  will  $  3.25  purchase  ? 

20.  If  a  train  of  cars  move  at  the  average  velocity  of  27m. 
3fur.  20rd.  per  lh.  20m.,  how  far  will  it  move  in  4h.  ? 

Ans.  82m.  2fur.  20rd. 


7mA 

I. 


PROPORTION.  23T 

PROPORTION. 

332.  A  proportion  is  an  equality  of  ratios.  Any  four 
numbers  are  in  proportion,  when  the  ratio  of  the  first  to  the 
second  is  the  same  as  that  of  the  third  to  the  fourth.  Thus,  the 
ratios  9  :  3  and  6  :  2,  being  equal  to  each  other,  when  written, 
9:3=  6  :  2,  or  §  =  § ,  form  a  proportion. 

Proportion  is  written  with  the  sign  of  equality  (==),  or,  as  is 
more  common,  with  four  dots  (: :),  between  the  ratios.  Thus, 
9:3=  6  :  2,  or  9  :  3  : :  6  :  2,  expresses  a  proportion,  and  is 
read,  The  ratio  of  9  to  3  is  equal  to  the  ratio  of  6  to  2,  or  9  is  to 

3  as  6  is  to  2. 

333.  The  terms  of  a  proportion  are  the  four  numbers  which 
form  the  proportion.  These  numbers  are  also  called  propor- 
tionals. The  first  and  third  terms,  or  proportionals,  are  called 
antecedents,  the  second  and  fourth  are  called  consequents  ;  the 
first  and  last  are  called  the  extremes,  the  second  and  third  the 
means  ;  the  first  and  second  compose  the  first  couplet,  the  third 
and  fourth  compose  the  second  ;  and  when  the  ratio  of  the  first 
of  three  terms  is  to  the  second  as  the  ratio  of  the  second  is  to 
the  third,  the  second  term  is  called  a  mean  'proportional  to  the 
other  two  terms. 

334.  A  direct  proportion  is  an  equality  between  two  direct 
ratios ;  an  inverse  or  reciprocal  proportion  is  an  equality  be- 
tween a  direct  and  an  inverse  or  reciprocal  ratio.  Thus,  the 
numbers  4,  2,  6,  3  are,  as  they  stand,  in  direct  proportion,  de- 
noting 4  :  2  : :  6  :  3  ;  but  in  the  order  4,  2,  3,  6,  are  in  inverse 
proportion,  denoting  that  4  :  2  : :  -£  :  £,  or  the  direct  ratio  of 

4  to  2  is  equal  to  the  inverse  ratio  of  3  to  6. 

Note.  —  The  term  proportion,  used  alone,  always  means  direct  proportion. 

335.  In  any  proportion,  if  the  antecedents  or  consequents,  or 
both,  are  divided,  or  multiplied,  by  the  same  number,  they  are 
still  proportionals.  Thus,  dividing  the  antecedents  of  the  pro- 
portion 4  :  8  : :  10  :  20  by  2,  we  have  2  :  8  : :  5  :  20  ;  dividing 
the  consequents  by  2,  we  have  4  :  4  : :  10  :  10 ;  and  dividing 
both  the  antecedents  and  consequents  by  2,  we  have  2:4:: 
5:10;  each  of  which  results  is  a  proportion,  since  if  we  divide 


238      I  PROPORTION. 

the  second  term  of  each  by  the  first,  and  the  fourth  by  the  third, 
the  two  quotients  will  be  equal.  The  effect  is  the  same  when 
the  terms  are  multiplied  by  the  same  number. 

336.  In  every  proportion  the  product  of  the  two  extremes  is 
equal  to  the  product  of  the  two  means.  Thus,  the  proportion 
16  :  8  ::  20  :  10  may  be  expressed  *£-  =  %%.  Now,  if  we  re- 
duce these  fractions  to  a  common  denominator,  we  have  *£g-  =* 
V6cr  5  Dut  m  this  operation  we  multiplied  together  the  two  ex- 
tremes of  the  proportion,  16  and  10,  and  the  two  means,  8  and 
20;  thus,  16  X  10  =  8  X  20.     Hence, 

1.  If  the  extremes  and  one  of  the  means  are  given,  the  other 
mean  may  he  found  by  dividing  the  product  of  the  extremes  by 
the  given  mean  ;  or, 

2.  If  the  means  and  one  of  the  extremes  are  given,  the  other 
extreme  may  be  found  by  dividing  the  product  of  the  means  by 
the  given  extreme. 

SIMPLE  PROPORTION. 

337.  Simple  Proportion  is  an  equality  between  two  simple 
ratios. 

Note.  —  Simple  Proportion  is  sometimes  called  the  Rule  of  Three,  and 
formerly  was  termed  by  arithmeticians  the  Golden  Rule. 

338.  The  object  of  that  part  of  simple  proportion  which  is 
usually  included  in  arithmetics,  is  to  find  a  fourth  proportional 
to  three  given  numbers,  or,  in  other  words,  to  find  the  fourth 
term  of  a  proportion,  when  the  other  three  terms  are  given. 

Ex.  1.  If  a  man  travel  243  miles  in  9  days,  how  far  will 
he  travel  in  24  days  ?  Ans.  648  miles. 

operation.  Since  9  days  have  the  same 

Extreme.    Mean.  Mean.         Extreme.       ratIo  tQ   24  d  ag   ^3   mileg> 

9  da.  :  2  4  da.  : :  2  4  3  m.  :  —  m.     the  distance  of  travel  in  9  days, 

2  4  have  to  the  distance  of  travel 

q  r~  £>  in  24  days,  we  have  the  first 

jo  /.  three   terms    of   a  proportion 

given,  namely,  the  two  means 

9)5832  and  one  of  the  extremes,  from 

which  to  find  the  required  ex- 

Ans.      6  4  8m.  Extreme,      treme.     Now,  to   arrange   the 

given  numbers  in  the  order  of 


PROPORTION.  239 

a  proportion,  or  state  the  question,  we  make  the  243  miles  the  third 
term,  because  it  is  of  the  same  kind  as  the  required  fourth  term,  and 
as  from  the  nature  of  the  question  the  latter  must  be  greater  than  the 
third  term,  we  make  the  greater  of  the  other  two  numbers  the  second 
term,  and  the  less  the  first;  and,  then,  the  product  of  the  means  di- 
vided by  the  given  extreme  gives  the  required  extreme  (Art.  336). 

By  Analysis.  —  If  a  man  travel  243  miles  in  9  days,  he  will  in 
1  day  travel  \  of  243  miles  =27  miles ;  then,  if  he  travel  27  miles 
in  1  day,  in  24  days  he  will  travel  24  times  27  miles  =  648  miles,  the 
answer,  as  before. 

By  Katio.  —  9  :  24  =  -fa  =  f ;  243  miles  -*-  f  =  648  miles,  Ans. 

2.  If  15  yards  of  cloth  cost  $  48.90,  what  will  5  yards  cost  ? 

Ans.  $16.30. 

operation.  "We   state  the   question   by   making 

1  5  :  5  : :  $  4  8.9  0  :  $  —     $  48.90  the  third  term,  because  it  is  of 

the  same  kind  as  the  required   term. 
$  X  4  8. 9  0  n  q  n         Then,  since  the  answer   must  be  less 

_ ==  1  6.3  0.        than  $  48.90,  because  5  yards  will  cost 

*■+*  less  than  15  yards,  we  make  5  yards, 

3         Ans.  $  16. oO.    ^he  less  of  the  two  numbers,  the  second 
term,  and  15  yards  the  first ;  and  pro- 
ceed as  in  the  first  example,  except  that  we  abridge  the  work  by  can- 
cellation. 

By  Analysis.  —  If  15  yards  cost  $  48.90,  1  yard  will  cost  -^  of 
$48.90  ==  $3.26;  then,  if  1  yard  cost  $3.26,  5  yards  will  cost  5 
times  $3.26  =$16.30. 

Rule.  —  Write  the  given  number  that  is  of  the  same  kind  as  the  re- 
quired fourth  term,  or  answer,  for  the  third  term  of  the  proportion. 

Of  the  other  two  numbers  write  the  larger  for  the  second  term,  and  the 
less  for  the  first,  when  the  answer  should  exceed  the  third  term ;  hut 
write  the  less  for  the  second  term,  and  the  larger  for  the  first,  when  the 
answer  should  be  less  than  the  third  term. 

Multiply  the  second  and  third  terms  together,  and  divide  their  product 
by  the  first;  or  divide  the  third  term  by  the  ratio  of  the  first  term  to  the 
second. 

Note  1.  —  When  the  first  and  second  terms  are  of  different  denominations, 
they  must  be  reduced  to  the  same  denomination ;  and  when  the  third  term  is 
a  compound  number,  it  must  be  reduced  to  the  lowest  denomination  men- 
tioned in  it.  The  answer  will  be  of  the  same  denomination  as  the  third 
term. 

Note  2.  — To  shorten  the  operations,  factors  common  to  the  dividend  and 
divisor  may  be  cancelled. 

Note  3. —  The  pupil  should  perform  these  questions  by  analysis,  as  well 
as  by  proportion,  and  introduce  cancellation  when  it  will  abbreviate  the  oper- 
ation. 


240  PROPORTION. 

Examples. 

3.  If  16  acres  of  land  cost  $  720,  what  will  197  acres 
cost  ?  Ans.  $  8865. 

4.  If  $  8865  buy  197  acres,  how  many  acres  may  be  bought 
for  $  720  ? 

5.  What  will  84hhd.  of  molasses  cost,  if  15hhd.  can  be  pur- 
chased for  $  175.95  ?  Ans.  $  985.32. 

6.  If  $  100  gain  $  6  in  12  months,  how  much  would  it  gain 
in  40  months  ?  Ans.  $  20. 

7.  If  a  certain  vessel  has  provisions  sufficient  to  last  a  crew 
of  10  men  45  days,  how  long  would  the  provisions  last  if  the 
vessel  were  to  ship  5  new  hands  ?  Ans.  30  days. 

8.  If  7  and  9  were  12,  what,  on  the  same  supposition,  would 
8  and  4  be  ? 

9.  If  9  men  can  perform  a  certain  piece  of  labor  in  17  days, 
how  long  would  it  take  3  men  to  do  it  ?  Ans.  51  days. 

10.  If  3  men  can  perform  a  piece  of  labor  in  51  days,  how 
many  must  be  added  to  the  number  to  perform  the  labor  in  17 
days  ?  Ans.  6. 

11.  A  rectangular  piece  of  land  containing  an  acre  is  5£  rods 
in  breadth.     What  is  its  length  ?  Ans.  29TJT  rods. 

12.  If  $100  gain  $  6  in  a  year,  how  much  will  $850 
gain  ? 

13.  If  $  100  gain  $  6  in  a  year,  how  much  would  be  suffi- 
cient to  gain  $  32  in  a  year?  Ans.  $  533.33^. 

14.  If  20  gallons  of  water  weigh  1671b.,  what  will  180  gal- 
lons weigh?  Ans.  15031b. 

15.  If  a  staff  3  feet  long  cast  a  shadow  of  2  feet,  how  high 
is  that  steeple  whose  shadow  is  75  feet?         Ans.  112 £  feet. 

16.  If  5fcwt.  be  carried  36  miles  for  $4.75,  how  far  might 
it  be  carried  for  $  160  ?  Ans.  1212|§  miles. 

17.  If  100  workmen  can  perform  a  piece  of  work  in  12 
days,  how  many  men  are  sufficient  to  perform  the  work  in  8 
days  ?  Ans.  150. 

18.  If  T72-  of  a  yard  cost  /^  of  a  dollar,  what  will  f  of  a  yard 
cost  ?  Ans.  $  0.48. 

19.  What  must  be  paid  for  21  A.  3R.  20p.  of  land,  if  36A. 
3R.  cost  $  1260  ?  Ans.  $  750. 


PROPORTION.  241 

20.  What  is  the  value  of  20001b.  of  standard  gold,  the  eagle, 
or  $  10  piece,  weighing  lOpwt.  18gr.  ? 

21.  If  4£  yards  of  cloth  cost  $9.75,  what  will  13£  yards 
cost  ?  Ans.  $  29.25. 

22.  "What  is  the  length  of  a  rectangle  whose  contents  are 
1  sq.  ft.  and  whose  breadth  is  2^  inches  ?      Ans.  57f  inches. 

23.  If  T7F  of  a  ship  cost  51£.,  what  are  -^  of  her  worth  ? 

Ans.  10£.  18s.  6fd. 

24.  If  the  moon  moves  13°  10'  35"  in  one  day,  in  what  time 
does  she  perform  one  revolution  ?        Ans.  27da.  7h.  43m. -\- 

25.  If  71b.  of  sugar  cost  f  of  a  dollar,  what  are  121b. 
worth?  Ans.  $  1.28f 

26.  If  $  1.75  will  buy  71b.  of  loaf-sugar,  how  much  will 
$  213.50  buy  ?  Ans.  8cwt.  2qr.  41b. 

27.  If  7  ounces  of  gold  are  worth  30£.,  what  i^;he  value  of 
71b.  lloz.  Ans.  407^2s.  lOfd. 

28.  A  friend  borrowed  of  me  $  500  for  6  months  ;  how  long 
ought  he  to  lend  me  $  600,  to  requite  the  favor  ? 

29.  If  the  penny  loaf  weighs  7oz.  when  flour  is  $  8  per 
barrel,  how  much  should  it  weigh  when  flour  is  $7.50  per 
barrel  ?  Ans.  7T\  ounces. 

30.  If  a  regiment  of  soldiers,  consisting  of  1000  men,  are  to 
be  clothed,  each  suit  to  contain  3  J  yards  of  cloth  that  is  1| 
yards  wide,  and  to  be  lined  with  flannel  1£  yards  wide,  how 
many  yards  will  it  take  to  line  the  whole  ?         Ans.  5625yd. 

31.  If  by  working  14  hours  per  day  C.  Simmons  can  plant 
half  of  a  field  in  9  days,  in  what  time  will  he  plant  the  remain- 
der, working  10  hours  per  day  at  the  same  rate  each  hour  ? 

Ans.  12f  days. 

32.  If  75  gallons  of  water  fall  into  a  cistern  containing  500 
gallons,  and  40  gallons  run  out,  in  an  hour,  in  what  time  will  it 
be  filled  ?  Ans.  14h.  17m.  8f  sec. 

33.  How  many  dozen  pairs  of  gloves,  at  $0.56  per  pair, 
can  be  bought  for  $  120.96  ?  Ans.  18  dozen. 

34.  A  certain  cistern  has  three  pipes ;  the  first  will  empty  it 
in  20  minutes,  the  second  in  40  minutes,  and  the  third  in  75 
minutes ;  in  what  time  would  they  all  «mpty  it  ? 

Ans.  11m.  19  if  sec. 

35.  A  can  mow  a  certain  field  in  5  days,  and  B  can  mow  it 

N   21 


242  PROPORTION. 

in  6  days ;  in  what  time  would  both  of  them  together  mow 
it  ?  Ans.  2T8T  days. 

36.  A  wall,  which  was  to  be  built  32  feet  high,  was  raised 
8  feet  by  6  men  in  12  days  ;  how  many  men  must  be  employed 
to  finish  the  wall  in  6  days  ? 

37.  A  can  build  a  boat  in  20  days,  but  with  the  assistance 
of  C  he  can  do  it  in  12  days ;  in  what  time  would  C  do  it 
alone  ?  Ans.  30  days. 

38.  In  a  fort  there  are  700  men  provided  with  1840001b.  of 
provisions,  of  which  each  man  consumes  51b.  a  week  ;  how  long 
can  they  subsist  ?  Ans.  52  weeks  4  days. 

39.  If  25  men  have  f-  of  a  pound  of  beef  each,  three  times  in 
a  week,  how  long  will  31501b.  last  them  ?       Ans.  56  weeks. 

40.  How  many  tiles  8  inches  square  will  lay  a  floor  20  feet 
long  and  lGrfeet  wide  ?  Ans.  720. 

41.  Ilovrmany  stones  10  inches  long,  9  inches  broad,  and  4 
inches  thick,  would  it  require  to  build  a  wall  80  feet  long,  20 
feet  high,  and  2£  feet  thick  ?  Ans.  17280  stones. 

42.  If  there  be  paid  for  1  ton  7cwt  3qr.  201b.  of  coal  $  9.50, 
what  will  13  tons  5cwt  2qr.  cost? 

43.  If  61.3  pounds  of  tea  cost  $44.9942,  what  is  the  price 
per  pound  ?  Ans.  $  0.734. 

44.  What  is  the  value  of  .15  of  a  hogshead  of  lime,  at  $  2.39 
per  hogshead  ?  Ans.  $  0.3585. 

45.  If  .75  of  a  ton  of  hay  cost  $  15,  what  is  it  per  ton  ? 

Ans.  $  20. 

46.  How  many  yards  of  carpeting  that  is  half  a  yard  wide 
will  cover  a  room  that  is  30  feet  long  and  1 8  feet  wide  ? 

Ans.  120  yards. 

47.  If  a  man  perform  a  journey  in  15  days,  when  a  day  is  12 
hours  long,  in  how  many  days  will  he  do  it  when  a  day  is  but 
10  hours  long? 

48.  If  450  men  are  in  a  garrison,  and  their  provisions  will 
last  them  but  5  months,  how  many  must  leave  the  garrison  that 
the  same  provisions  may  be  sufficient  to  supply  the  remaining 
men  9  months  ?  Ans.  2(|p  men. 

49.  The  hour  and  minute  hands  of  a  watch  are  together  at 
12  o'clock  ;  when  will  they  next  be  together  ? 

Ans.  lh.  5m.  27T\see. 


PROPORTION.  243 

50.  A  and  B  can  perform  a  piece  of  work  in  5T5T  days,  B 
and  C  in  6§  days,  and  A  and  C  in  6  days  ;  in  what  time  would 
each  of  them  perform  the  work  alone,  and  how  long  would  it 
take  them  to  do  the  work  together  ? 

Ans.  A  would  do  the  work  in  10  days ;  B,  in  12  days ;  C, 
in  15  days ;  A,  B,  and  C  together,  in  4  days. 

339.  To  divide  a  number  or  quantity  into  parts,  which  are 
proportional  to  given  numbers. 

Ex.  1.  Divide  $250  into  two  parts  which  shall  be  one  to 
the  other  as  2  to  3.  Ans.  '$  100  and  $  150. 

operation.  Since  the  parts 

2  -\-  3  =  5  are  to  be  propor- 

5  :  3  :  :  $  250  :  $  150,  the  greater  part, )   A        t[°™1  to  2  and  3> 

5  :  2  :  :  $  250  :  $  100,  the  less  part,        f  Ans*  7[h?se  g*  »  5> 

r  J  it  is  evident  that 

the  sum  of  the 
two  numbers,  5,  will  have  the  same  ratio  to  the  greater  of  them,  3,  as 
the  amount  to  be  divided,  $  250,  has  to  the  greater  of  the  required 
parts ;  and  that  the  sum,  5,  will  have  the  same  ratio  to  the  less  num- 
ber, 2,  as  the  $  250  has  to  the  less  of  the  required  parts  ;  we  there- 
fore make  two  statements,  and  then  find  the  required  term  of  each 
proportion  as  in  Art.  337.     Hence, 

As  the  sum  of  the  given  numbers  is  to  any  one  of  them,  so  is  the  whole 
quantity  to  be  divided  to  the  part  corresponding  to  the  number  used  as 
the  second  term. 

Note.  —  This  application  of  proportion  is  sometimes  called  Distributive  or 
Partitive  Proportion. 

Examples. 

2.  A  farmer  divides  between  his  three  sons  246A.  IK.  32p. 
of  land,  sharing  it  between  them  as  the  numbers  3,  4,  and  5. 
What  were  the  shares  ? 

Ans.  61A.  2R.  18p. ;  82A.  0E.  24p. ;  102A.  2R.  30p. 

3.  Divide  319  into  four  parts,  that  shall  be  to  each  other  as 
the  numbers  4£,  6£,  6-f ,  and  7. 

Ans.55ff£;  85¥yT;  86f  f f  ;  91fff. 

4.  Standard  gold  for  coinage  consists  of  9  parts  of  pure  gold, 
and  1  part  alloy.  Allowing  the  alloy  to  be  silver  and  copper 
in  equal  parts,  how  much  pure  gold,  silver,  and  copper  are  con- 
tained in  a  double  eagle,  its  weight  being  loz.  lpwt.  12gr.  ? 

Ans.  19pwt.  8fgr.  gold;  lpwt.  lfgr.  silver;  lpwt.  lfgr. 
copper. 


244  PROPORTION. 

5.  The  half-dollar  of  the  United  States  coinage  weighs  192 
grains  Troy,  and  consists  of  9  parts  pure  silver  and  1  part  of 
copper.  How  much  pure  silver  and  how  much  copper  in  20 
half-dollars?  Ans.  7oz.  4pwt.  silver;  16pwt.  copper. 

6.  Divide  $  600  between  three  men,  so  that  the  second  man 
shall  receive  one  third  more  than  the  first,  and  the  third  man 
shall  receive  two  thirds  more  than  the  second. 

7.  A,  B,  and  C  freight  a  steamer ;  A  puts  on  board  98  tons, 
B  86  tons,  and  C  64  tons.  Owing  to  danger  of  being  wrecked, 
there  were  thrown  overboard  while  at  sea  93  tons.  "What 
should  be  the  number  of  tons  lost  by  each  ? 

Ans.  A,  36 j  tons  ;  B,  32£  tons  ;  and  C,  24  tons. 

8.  A  and  B  start  together  by  railroad  from  Chicago  for  Ga- 
lena; A  travels  by  freight  train  at  the  rate  of  15  miles  per 
hour,  and  B  by  passenger  train  at  the  rate  of  25  miles  per 
hour.  C  leaves  Galena  for  Chicago  at  the  same  time  by  ex- 
press train,  whose  velocity  is  at  the  rate  of  32  miles  per  hour. 
Allowing  the  distance  between  the  two  places  to  be  160  miles, 
how  far  from  Chicago  will  A  and  B  each  be,  when  C  passes 
them  ?  Ans.  A,  51^  miles  ;  B  70-^  miles. 

COMPOUND  PROPORTION. 

340.  A  Compound  Proportion  is  an  expression  of  equality 
between  a  compound  and  a  simple  ratio.     Thus, 

4*3) 

k  [  „  V  :  :  60  :  63,  is  a  compound  proportion. 

Compound  proportion  is  employed  in  the  solution  of  such 
questions  as  would  require  two  or  more  statements  in  Simple 
Proportion. 

Ex.  1.  If  8  men  spend  $  32  in  13  weeks,  what  will  24  men 
spend  in  52  weeks  ?  Ans.  $  384. 

operation.  In   stating   the 

Extreme.  Mean.  question,  we  make 

8                     a  a                    \          Mean.  Extreme.              1  „  ~        ■  •  1     •        t* 

men     :  24  men       I .  .  <&  30  •  <fc ®     '  wmcn  1S  of 

13  weeks  :  52  weeks    j* '  *  the  same  kind  as 

o               a  the  required  term, 

J  .          .  *    ,    _ .  the     third    term. 

3  4  X  g  ff  X  3  2  Then,takingofthe 

*&  v   r  *A  S*  *               remaining    terms 

*  A  A  *  two  of  the  same 


PKOPORTION.  245 

kind,  8  men  and  24  men,  we  inquire  whether  the  answer  depending 
on  these  alone  must  be  greater  or  less  than  the  third  term ;  and  since 
it  must  be  greater,  because  24  men  will  spend  more  than  8  men  in 
the  same  time,  we  make  24  men  the  second  term,  and  8  men  the  first 
term.  Again,  we  take  the  two  remaining  terms,  and  make  52  weeks 
the  second  term,  and  13  weeks  the  first,  since  the  same  number  of 
men  would  spend  more  in  52  weeks  than  in  13  weeks.  We  then  find 
the  continued  product  of  the  second  and  third  terms,  and  divide  by 
the  product  of  the  first  terms. 

By  Analysis. — If  8  men  spend  $32  in  13  weeks,  1  man  will 
spend  $-3g2-  =  $4  in  13  weeks,  and  24  men  will  spend  $4  X  24  = 
$  96  in  13  weeks.  If  24  men  spend  $96  in  13  weeks,  in  1  week 
they  will  spend  $ff,  and  in  52  weeks,  $ff  X  52  =  $  384,  Ans. 

By  Ratio.  —  The  ratio  of  8  :  24  =  J  ;  the  ratio  of  13  :  52  =  £  ; 
S32-T-1  x  £  =  $384,  Ans. 

Note.  —  To  have  solved  the  question  by  simple  proportion,  two  statements 
would  have  been  required,  which  would  have  produced  the  following  pro- 
portions :  — 

8  men      :  24  men      :  :  $32  :  $96. 
13  weeks  :  52  weeks  :  :  $96  :  $384. 

Rule.  —  Make  that  number  which  is  of  the  same  kind  as  the  answer 
required  the  third  term  of  a  proportion.  Of  the  remaining  numbers, 
take  any  two,  that  are  of  the  same  kind,  and  consider  whether  an  answer 
depending  upon  these  alone  would  be  greater  or  less  than  the  third  term, 
and  place  them  as  directed  in  Simple  Proportion. 

Then  take  any  other  two,  and  consider  whether  an  answer  depending 
only  upon  them  would  be  greater  or  less  than  the  third  term,  and  ar- 
range them  accordingly ;  and  so  on  until  all  are  used. 

Multiply  the  product  of  the  second  terms  by  the  third,  and  divide  the 
result  by  the  product  of  the  first  terms.  The  quotient  will  be  the  fourth 
term,  or  answer. 

Examples. 

2.  If  a  man  can  travel  117  miles  in  30  days,  travelling  9 
hours  a  day,  how  far  can  he  go  in  20  days,  travelling  12  hours 
a  day  ?  Ans.  104  miles. 

3.  If  6  men  in  16  days  of  9  hours  each  build  a  wall  20  feet 
long,  6  feet  high,  and  4  feet  thick,  in  how  many  days  of  8  hours 
each  will  24  men  build  a  wall  200  feet  long,  8  feet  high,  and  6 
feet  thick  ? 

4.  If  $  100  gain  $  6  in  one  year,  how  much  would  $  500 
gain  in  four  months  ?  Ans.  $  10. 

5.  If  $  100  gain  $  6  in  one  year,  what  must  be  the  sum  to 
gain  $  10  in  4  months  ?  Ans.  $  500. 

21* 


246  PROPORTION. 

6.  How  long  will  it  take  $  500  to  gain  $  10,  if  $  100  gain 
$  G  in  one  year  ?  Ans.  4  months. 

7.  If  $  500  gain  $  10  in  4  months,  what  is  the  rate  per  cent  ? 

8.  If  5  compositors  in  16  days,  11  hours  long,  can  compose 
25  sheets  of  24  pages  in  each  sheet,  44  lines  in  a  page,  and 
40  letters  in  a  line,  in  how  many  days,  10  hours  long,  can  9 
compositors  compose  a  volume  (to  be  printed  on  the  same  kind 
of  type),  consisting  of  36  sheets,  16  pages  to  a  sheet,  50  lines 
to  a  page,  and  45  letters  in  a  line  ?  Ans.  12  days. 

9.  If  12  men  can  build  a  wall  30  feet  long,  6  feet  high,  and 
3  feet  thick,  in  15  days,  when  the  days  are  12  hours  long,  in 
what  time  will  60  men  build  a  wall  300  feet  long,  8  feet  high, 
and  6  feet  thick,  when  they  work  only  8  hours  a  day  ?       \  f 

Ans.  120  days. 

10.  If  16  horses  consume  84  bushels  of  grain  in  24  days, 
how  many  bushels  will  suffice  32  horses  48  days  ? 

Ans.  336  bushels. 

11.  If  the  carriage  of  5cwt.  3qr.  150  miles  cost  §24.58, 
what  must  be  paid  for  the  carriage  of  7cwt.  3qr.  64  miles,  at 
the  same  rate  ?  Ans.  $  14.135—]—. 

12.  If  7£oz.  of  bread  be  bought  for  4£d.  when  corn  is  4s.  2d. 
per  bushel,  what  weight  of  it  may  be  bought  for  Is.  2d.  when 
the  price  per  bushel  is  5s.  6d.  ? 

13.  If  496  men,  in  5£  days  of  11  hours  each,  dig  a  trench 
of  7  degrees  of  hardness,  465  feet  long,  3§  wide,  2£  deep,  in 
how  many  days  of  9  hours  long  will  24  men  dig  a  trench  of  4 
degrees  of  hardness,  337£  feet  long,  5f  wide,  and  3£  deep  ? 

Ans.  132  days. 

CONJOINED  PROPORTION. 

341.  A  Conjoined  Proportion  is  a  proportion  which  has 
each  antecedent  of  a  compound  ratio  equal  in  value  to  its  con- 
sequent. 

The  first  term  of  each  pair  of  equivalent  terms  is  an  ante- 
cedent, and  the  term  following  a  consequent. 

Note.  —  This  kind  of  proportion  is  sometimes  called  the  Chain  Rule. 
Ex.  1.     If  10  barrels  of  apples  will  pay  for  5  cords  of  wood, 


PROPORTION.  247 

and  20  cords  of  wood  for  4  tons  of  hay,  how  many  barrels  of 
apples  will  it  take  to  purchase  50  tons  of  hay  ?       Ans.  500. 

operation.  We  arrange  the  an- 

Antecedents.  Consequents.  tecedents  at  the  left 

1  0  barrels  =*=  5  cords.  ?/  the  W  <*  e^al" 
~  ~         t              4   ;                                 lty,  and    the   conse- 

2  0  cords     *=  4  tons.  quents  at  the  rf  ht  of 

5  0  tons       =  —  barrels.  tne  same ;  and,  as  the 

10  X  f&0  X  50  answer  is  to  be  of  the 

z=z  5  0  0  barrels,  Ans.   same  kind  as  the  first 

j£  X  4  term,we  place  the  odd 

term  in  the  column  of 
antecedents.  Then,  to  find  the  term  corresponding  in  value  with  the 
odd  term,  we  divide  the  continued  product  of  the  10,  20,  and  50  of 
the  column  of  antecedents,  by  the  product  of  the  5  and  4  of  the 
column  of  consequents  ;  and  obtain  500  barrels,  Ans. 

By  Analysis.  —  If  4  tons  of  hay  equal  in  value  20  cords  of  wood, 
1  ton  equals  in  value  5  cords  of  wood.  If  5  cords  of  wood  equal  10 
barrels  of  apples,  then  1  ton  of  hay  equals  10  barrels  of  apples,  and 
50  tons  equal  50  times  the  value  of  10  barrels  of  apples,  or  500  bar- 
rels, answer  as  before. 

Rule.  —  Place  the  antecedents  in  one  column  and  the  consequents 
in  another ',  on  the  right,  with  the  sign  of  equality  between  them.  Divide 
the  continued  product  of  the  terms  in  the  column  containing  the  odd  term 
by  the  continued  product  of  the  other  column,  and  the  quotient  will  be 
the  answer. 

Note  1.  —  The  odd  term  will  always  belong  to  that  column  which  does  not 
contain  the  other  term  of  the  same  kind. 

Note  2.  —  Conjoined  proportion  is  nothing  else  than  simple  or  compound 
proportion,  exhibited  in  a  different,  and  in  many  cases  a  more  convenient 
form ;  and,  to  show  the  correctness  of  the  rule,  the  following  examples  may 
also  be  performed  by  either  simple  or  compound  proportion. 

Examples. 

2.  If  100  acres  in  Bradford  be  worth  120  in  Haverhill,  and 
50  in  Haverhill  be  worth  65  in  Methuen,  how  many  acres  in 
Bradford  are  equal  to  150  in  Methuen  ?       Ans.  96f^  acres. 

3.  If  101b.  of  cheese  are  equal  in  value  to  71b.  of  butter,  and 
111b.  of  butter  to  2  bushels -of  corn,  and  11  bushels  of  corn  to 
8  bushels  of  rye,  and  4  bushels  of  rye  to  one  cord  of  wood, 
how  many  pounds  of  cheese  are  equal  in  value  to  10  cords  of 
wood?  Ans.  432^1b. 

4.  If  12  men  can  do  as  much  work  as  25  women,  and  5 


248  PROPORTION. 

women  do  as  much  as  6  boys,  how  many  men  would  it  take  to 
do  the  work  of  75  boys  ?  Ans.  30  men. 

5.  If  6  gallons  liquid  measure  equal  5  English  imperial  gal- 
lons, and  if  10  English  imperial  gallons  equal  6  velts  of 
France,  and  if  26  velts  of  France  equal  16  Russian  vedros  of 
wine,  how  many  vedros  of  wine  equal  63  gallons  liquid  meas- 
ure ? 

6.  If  7  bushels  of  corn  in  Boston  be  worth  8  bushels  in  Buf- 
falo, and  10  bushels  in  Buffalo  be  worth  14  bushels  in  Chicago, 
and  21  bushels  in  Chicago  be  worth  25  bushels  at  Davenport, 
how  many  bushels  in  Boston  are  worth  1200  bushels  at  Daven- 
port ?  Ans.  630  bushels. 

7.  If  24  shillings  in  Massachusetts  are  equal  to  32  shillings 
in  New  York,  and  if  48  shillings  in  New  York  are  equal  to  45 
shillings  in  Pennsylvania,  and  if  15  shillings  in  Pennsylvania 
are  equal  to  10  shillings  in  Canada,  how  many  shillings  in  Can- 
nada  are  equal  to  100  shillings  in  Massachusetts  ? 

Ans.  83£  shillings. 

MISCELLANEOUS  EXAMPLES  IN  PROPORTION. 

1.  A  sets  out  on  a  journey,  and  travels  27  miles  a  day ;  7 
days  after,  B  sets  out  and  travels  the  same  road  36  miles  a  day ; 
in  how  many  days  will  B  overtake  A?  Ans.  21  days. 

2.  If  I  sell  coffee  at  2s.  3d.  per  pound  and  gain  35  per  cent, 
what  did  I  give  per  pound  ? 

3.  A  detachment  of  2000  soldiers  were  supplied  with  bread 
sufficient  to  last  them  12  weeks,  allowing  each  man  14  ounces 
a  day;  but  on  examination  they  find  105  barrels,  containing 
2001b.  each,  wholly  spoiled ;  how  much  a  day  may  each  man 
eat,  that  the  remainder  may  supply  them  12  weeks? 

Ans.  12oz. 

4.  In  consequence  of  having  a  seventh  part  of  their  bread 
spoiled,  2000  soldiers  were  put  on  an  allowance  of  12  ounces  of 
bread  per  day  for  12  weeks ;  what  was  the  whole  weight  of  their 
bread  (good  and  bad),  and  how  much  was  spoiled  ? 

Ans.  The  whole  weight,  1470001b.;  spoiled,  210001b. 

5.  Two  thousand  soldiers,  having  lost  105  barrels  of  bread, 
weighing  2001b.  each,  were  obliged  to  subsist  on  12  ounces  a 


PROPORTION .  249 

day  for  12  weeks ;  but  had  none  been  lost,  they  might  have 
had  14  ounces  a  day  for  the  same  time.  What  was  the  whole 
weight,  including  what  was  lost,  and  how  much  had  they  left  to 
subsist  on  ? 

Ans.    The   whole   weight,    1470001b.;    left   to   subsist   on, 
1260001b. 

6.  Bought  threescore  pieces  of  Hollands  for  three  times  as 
many  dollars,  and  sold  them  again  for  four  times  as  many  dol- 
lars ;  but  if  they  had  cost  me  as  much  as  I  sold  them  for,  for 
what  should  I  have  sold  them  to  gain  at  the  same  rate  ? 

Ans.  $  320. 

7.  Bought  20  pounds  of  tea  at  the  rate  of  1J-  pounds  for  a 
dollar,  and  62  pounds  more  at  the  rate  of  15  pounds  for  $  12 ; 
and  sold  the  whole  at  the  rate  of  3  pounds  for  $  4.  How  much 
did  I  gain,  or  lose  ? 

8.  Each  of  the  four  sides  of  a  certain  field  appeared  to  be  2 
furlongs  30  rods  and  3  yards  in  length  when  measured  by  a  line 
supposed  to  be  4  rods  long ;  but  the  line  was  found  to  have 
been  only  64  feet  in  length.  Required  the  true  distance  round 
the  field.  Ans.  7074^  feet. 

9.  If  5  oxen  or  7  cows  eat  3T4T  tons  of  hay  in  87  days,  in 
what  time  will  2  oxen  and  3  cows  eat  the  same  quantity  of 
hay?  Ans.  105  days. 

10.  If  360  men  be  placed  in  a  garrison,  and  have  provisions 
for  6  months,  how  many  men  must  be  sent  away  at  the  end  of 
4  months  that  the  remaining  provisions  may  last  the  garrison  8 
months  longer  ?  Ans.  270  men. 

11.  My  tailor  informs  me  it  will  take  10£  square  yards 
of  cloth  to  make  me  a  full  suit  of  clothes.  The  cloth  I 
am  about  to  purchase  is  1|  yards  wide,  and  on  sponging  it 
will  shrink  J$  in  width  and  length.  How  many  yards  of  the 
above  cloth  must  I  purchase  for  my  "  new  suit    1    * 

Ans.  6T§f§-yd. 

12.  If  the  price  of  a  farm  of  130A.  2R.  20p.  be  $  6537.50, 
what  will  be  the  price  of  another,  containing  100 A.  0E.  30p., 
if  4  acres  of  the  latter  be  worth  5  of  the  former  ? 

Ans.  $6267.71+. 

13.  How  much  each  of  copper,  tin,  zinc,  and  lead  will  be 
required  to  make  a  bell  weighing  192  tons  17cwt.  161b.,  the 


250  PROPORTION. 

weight  of  that  at  Moscow,  bell-metal  being  composed  of  800 
parts  copper,  101  tin,  56  zinc,  and  43  lead? 

Ans.  Copper,308572|lb.;  tin,  38957^1b.;  zinc,  21600iSlb.; 
lead,  16585^-f  Jib. 

14.  The  relative  heating  power  per  cord  of  white-oak  to 
white-pine  being  as  81  to  42,  and  that  of  white-pine  to  pitch- 
pine  as  7  to  8 ;  how  many  cords  composed  of  white-pine  and 
pitch-pine  in  equal  parts  will  be  required  to  produce  as  much 
heat  as  20  cords  of  white-oak?  Ans.  18  cords  of  each. 

15.  If  a  cask  of  wine,  containing  85  gallons,  cost  $116.95, 
what  would  be  the  value  of  a  hogshead  containing  63  gallons, 
and  composed  of  four  parts  of  the  same  wine  and  one  part  of 
water?  Ans.  $  69.34f§. 

16.  If  8  horses  in  30  days  consume  3£  tons  of  hay,  how  long 
will  4^  tons  last  10  horses,  15  cows,  and  7  sheep,  each  cow 
consuming  f-  as  much  as  a  horse,  and  each  sheep  ^  as  much  as 
a  cow  ? 

17.  The  relative  nutritious  matter  per  pound  of  potatoes  to 
beets  being  as  25  to  14,  that  of  beets  to  carrots  as  7  to  5,  and 
that  of  carrots  to  turnips  as  5  to  2 ;  how  many  pounds  of  turnips 
are  equal  in  nutritious  matter  to  2000  pounds  of  potatoes?  and 
allowing  60  pounds  to  a  bushel,  how  much  cheaper  or  dearer 
relatively  per  bushel  are  potatoes  at  80  cents,  and  turnips  at  20 
cents,  than  either  beets  at  35  cents  or  carrots  at  25  cents. 

'Ans.  12500  pounds;  potatoes  17 Jets,  dearer;  turnips  lOcts. 
dearer. 

18.  A  and  B  set  out  from  the  same  place,  and  in  the  same 
direction.  A  travels  uniformly  18  miles  per  day,  and  after  9 
days  turns  and  goes  back  as  far  asB  has  travelled  during  those 
9  days ;  he  then  turns  again,  and,  pursuing  his  journey,  over- 
takes B  22  J  days  after  the  time  they  first  set  out.  It  is  re- 
quired to  find  the  rate  at  which  B  uniformly  travelled. 

Ans.  10  miles  per  day. 

19.  Two  men  in  Boston  hire  a  carriage  for  $  25,  to  go  to  Con- 
cord, N.  H.,  and  back,  the  distance  being  72  miles,  with  the  privi- 
lege of  taking  in  three  more  persons.  Having  gone  20  miles, 
they  take  in  A ;  at  Concord,  they  take  in  B  ;  and  when  within  30 
miles  of  Boston,  they  take  in  C.    How  much  shall  each  man  pay  ? 

Ans.  First  man,  $7.G09$$f ;  second  man,  $  7.609  f$f;  A 
$£.873^;  B,$2.86417J;  C.  $1,041^. 


PERCENTAGE. 


251 


PERCENTAGE. 


342.  Percentage  is  an  allowance  for  every  hundred,  con- 
sisting of  a  part  of  that  on  which  it  is  reckoned. 

Per  cent.,  a  contraction  of  per  centum,  signifies  by  the  hun- 
dred. By  1  per  cent,  is  to  be  understood  1  out  of  a  hundred, 
or  1  hundredth  ;  by  2  per  cent,  is  to  be  understood  2  out  of  a 
hundred ;  or  2  hundredths,  &c. 

343 o  The  rate  per  cent,  is  the  number  denoting  the  part  al- 
lowed for  every  hundred,  as  5  per  cent.,  6  per  cent.,  &c. 

344.  The  basis  of  percentage  is  the  number  on  which  per- 
centage is  reckoned. 

345.  The  amount  is  the  basis  of  percentage  increased  by 
the  percentage ;  and  the  remainder  is  the  basis  decreased  by 
the  percentage. 

346.  The  rates  per  cent,  may  be  written  decimally  as  in  the 
following 

Table. 


1  per  cent. 

= 

.01 

9  per  cent.  = 

.09 

2  per  cent. 

a 

.02     . 

10  per  cent.    ' 

.10 

3  per  cent. 

u 

.03 

15  per  cent.   * 

.15 

4  per  cent. 

a 

.04 

20  per  cent.   (< 

.20 

5  per  cent. 

u 

.05 

25  per  cent.    i 

.25 

6  per  cent. 

a 

.06 

63  per  cent.   " 

.63 

7  per  cent. 

a 

.07 

100  per  cent.   * 

1.00 

8  per  cent. 

a 

.08 

116  per  cent.   ' 

1.16 

And 

\  per  cent,  may  be  written  .005,  or  .00^,  and  be  read  -fa  of  1  per 
cent,  or  ^  of  1  per  cent. 

\  per  cent,  may  be  written  .0025,  or  .00 J,  and  be  read  -gfe  of  1 
per  cent.,  or  \  of  1  per  cent. 

\  per  cent,  may  be  written  .00125,  or  .00i,  and  be  read  ijtffo  of 
1  per  cent,  or  ^  of  1  per  cent. 

1\  per  cent,  may  be  written  .075,  or  .07^,  and  be  read  7^  per 
cent.,  or  7-|  per  cent. 

Note.  —  If  a  fraction  of  1  per  cent,  cannot  be  exactly  expressed  in  decimal 
figures,  it  may  be  written  as  a  part  of  a  mixed  decimal.  Thus,  4i  per  cent, 
may  be  written  .04g;  67  per  cent,  may  be  written  .061;  and  llg  per  cent. 
may  be  written  .11?. 


252  PERCENTAGE. 

Exercises. 

1.  Express  decimally  19  per  cent!  Ans.  .19. 

2.  Express  decimally  27  per  cent. 

3.  Express  decimally  13^  per  cent. 

4.  Express  decimally  If  per  cent. 

5.  Express  decimally  7f  per  cent. 

6.  Express  decimally  77£  per  cent. 

7.  Express  decimally  106  per  cent. 

8.  Express  decimally  107  per  cent. 

9.  Express  decimally  305  per  cent. 
10.  Express  decimally  999 J-f  per  cent. 

347.  To  find  the  percentage  any  given  rate  per  cent,  is  of 
any  number  or  quantity. 

Ex.  1.  Bought  £  of  a  ship  for  $  15650,  and  sold  the  same 
at  a  gain  of  12  per  cent.  How  much  did  I  make  by  the  trans- 
action? Ans.  $1878. 

operation.  Since  12  per  Cent.  equals 

Basis  of  percentage,  $  1  5  6  5  0  12  of  the  ori„inal  cost\  we 

Eate  per  cent.         A2  multiply  $15650    by  the 

Percentage,             $  1  8  7  8.0  0  Ans.  decimal  expression  .12. 

Rule.  — Multiply  the  given  number  by  the  rate  per  cent,  expressed 
decimally,  and  the  product  will  be  the  percentage.     Or, 

As  100  per  cent,  is  to  the  given  rate  per  cent.,  so  is  the  given  basis  of 
percentage  to  the  percentage  required. 


2.  What 

3.  What 

4.  What 

5.  What 

6.  What 

7.  What 

8.  What 

9.  What 

10.  What 

11.  What 

12.  What 


Examples. 

s  15  per  cent,  of  500  bushels  ?  Ans.  75bu. 

s  20  per  cent,  of  75cwt.  ?  Ans.  15cwt. 

s  30  per  cent,  of  150  tons  ?  Ans.  45  tons. 

s  75  per  cent,  of  $  500  ? 

s  95  per  cent,  of  700  chaldrons  ? 

s  2  per  cent,  of  40  miles  ?  Ans.  .8  mile. 

s  99  per  cent,  of  $  1000  ?  Ans.  $  990. 

s  33£  per  cent,  of  144  barrels  ?       Ans.  48bbl. 

s  66§  per  cent,  of  90  hogsheads  ? 

s  i  per  cent,  of  $  100  ?  Ans.  $  0.25. 

s  J  per  cent,  of  17281b.  ?  Ans.  15.121b. 


13.  A  certain  colonel,  whose  regiment  consisted  of  900  men, 


PERCENTAGE.  253 

lost  8  per  cent,  of  them  in  battle,  and  50  per  cent,  of  the  re- 
mainder by  sickness.     How  many  had  he  remaining  ? 

Ans.  414  men. 

14.  A  merchant,  having  $  1728  in  the  Union  Bank,  wishes 
to  withdraw  15  per  cent. ;  how  much  will  remain  ? 

15.  A  gentleman,  who  had  an  estate  of  $  25,000,  in  his  will 
gave  to  his  wife  40  per  cent,  of  his  property,  and  to  his  son 
Samuel  30  per  cent,  of  the  remainder.  The  residue  he  di- 
vided equally  among  his  daughters,  Marcia,  Isabella,  and  Clara, 
after  having  deducted  $  60  as  a  present  to  his  clergyman. 
What  did  each  receive  ? 

Ans.  Wife,  $  10,000 ;  son,  $  4,500 ;  daughters,  $  3,480  each. 

348 #  To  find  what  rate  per  cent,  one  given  number  is  of 
another. 

Ex.  1.     What  per  cent,  of  50  is  12  ?       Ans.  24  per  cent. 

operation.  Since  the  percentage  equals 

IS  —  A  —  -24?  Ans.  the  product  of  the  basis  of 

2  percentage  by  the  number 

12  X  X  Vl  Cl  denoting  the  rate  per  cent. 

Qr    zJl  —  24  per  cent.       (4r.L    343)>   the  ^  quotient 

ft  0  arising   from    dividing    the 

percentage  by  the  number 
denoting  the  basis  must  equal  the  rate  per  cent.  We  therefore 
divide  12  by  the  50,  and  obtain  \%  =  -^,  which,  expressed  decimally, 
equals  .24,  or  24  per  cent.  Since  the  question,  evidently,  is  the 
same  as  to  find  if  of  100  per  cent,  we  multiply  the  12  by  100,  or 
annex  two  ciphers  and  divide  by  50,  and  obtain  the  same  result  as 
before. 

Rule.  — Annex  two  ciphers  to  the  number  denoting  the  percentage, 
and  divide  by  the  number  on  which  the  percentage  is  reckoned  ;  and  the 
quotient  will  be  the  rate  per  cent.     Or, 

As  the  given  basis  of  percentage  is  to  the  given  percentage,  so  is  100 
per  cent,  to  the  rate  per  cent,  required. 

Examples. 

2.  What  per  cent,  of  16  is  2  ?  Ans.  12J  per  cent. 

3.  What  per  cent,  of  110  is  11  ?  Ans.  10  per  cent. 

4.  What  per  cent,  of  2£  is  £  ? 

5.  £  of  18  per  cent,  is  what  per  cent,  of  24  per  cent.  ? 

Ans.  25  per  cent. 

6.  What  per  cent,  of  $  150  is  25  per  cent,  of  $  36  ? 

N      22  Ans.  6  per  cent. 


254  PERCENTAGE. 

7.  36  bushels  is  what  per  cent,  of  48  bushels  ? 

Ans.  75  per  cent. 

8.  What  per  cent,  of  4  years  is  1  year  6  months  ? 

9.  31  gallons  2  quarts  is  what  per  cent,  of  1  hogshead  ? 

Ans.  50  per  cent. 

10.  Of  160  yards  of  cloth  there  have  been  sold  128  yards ; 
what  per  cent,  of  the  whole  remains  unsold  ? 

Ans.  20  per  cent. 

11.  What  per  cent,  of  J-  of  f  of  §  is  -J  ?    Ans.  20  per  cent. 

12.  In  a  certain  school  the  number  of  pupils  studying  ge- 
ography is  40  per  cent,  more  than  the  number  studying  gram- 
mar. What  per  cent,  less  is  the  number  studying  grammar 
than  the  number  studying  geography  ?       Ans.  28f-  per  cent. 

13.  If  a  miller  takes  out  4  quarts  toll  from  every  bushel  he 
grinds,  what  per  cent,  does  he  take  for  toll  ? 

14.  If  a  certain  coin  is  made  of  22  parts  copper  and  3  parts 
nickel,  what  per  cent,  of  it  is  copper,  and  Avhat  per  cent, 
nickel?  Ans.  88  per  cent,  copper  ;  12  per  cent,  nickel. 

15.  In  a  certain  orchard  there  are  250  trees,  of  which  40 
per  cent,  are  apple-trees,  12  per  cent,  cherry-trees,  8  per  cent, 
plum-trees,  and  the  remainder,  with  the  exception  of  25  pear- 
trees,  consist  of  peach-trees.  What  per  cent,  of  the  whole  are 
the  peach-trees  ?  Ans.  30  per  cent. 

349t  To  find  a  number  when  a  given  number  is  known  to 
be  a  certain  per  cent,  of  it. 

Ex.  1.  I  have  bought  two  house-lots ;  for  the  one  I  paid 
$  300,  which  was  60  per  cent,  of  what  I  paid  for  the  other. 
What  did  I  pay  for  the  latter  ?  Ans.  $  500. 

operation.  Since  $300 

$300  -r-  60  =  $5  ;  $5  X  100  =  $500,  Ans.  is  60  per  cent. 

5  of  the  unknown 

*  w^        ^,-v  sum>     1      Per 

0r    W0X1OO_S5OO.  cent,      of     it 

00  must  equal  fo 

w  of    $300,    or 

$  5 ;  and  100  per  cent.,  or  the  whole  of  it,  must  equal  100  times  $  5, 

or  $  500,  Ans.     And  since  the  question  is  evidently  the  same  as  to 

find  the  value  of  ^°/  of  $  300,  we  multiply  the  $  300  by  100,  and 

divide  by  60,  and  obtain  the  same  result  as  before. 

Rule.  —  Annex  two  ciphers  to  the  number  denoting  the  percentage, 
and  divide  by  that  denoting  the-  rate  per  cent.      Or, 


PERCENTAGE.  255 

As  100  per  cent,  is  to  the  given  rate  per  cent.,  so  is  the  given  per- 
centage to  the  basis  of  percentage  required. 

Examples. 

2.  25  is  10  per  cent,  of  what  number?  Ans.  250. 

3.  16£  is  8  per  cent,  of  what  number  ?  Ans.  203|. 

4.  72  is  12  per  cent,  of  what  number  ? 

5.  J  is  40  per  cent,  of  what  number  ?  Ans.  2T^-. 

6.  $1.62£  is  12 i  per  cent,  of  how  many  dollars  ?  Ans.  $  13. 

7.  $  1^  is  £  per  cent,  of  what  sum  ?  Ans.  $  140. 

8.  A  flock  of  sheep  has  lost  15-J-  per  cent,  of  its  number, 
17  sheep  having  been  killed  by  the  dogs,  and  6  more  having 
been  drowned.     What  was  the  original  number  ?    Ans.  150. 

9.  If  a  man  owning  45  per  cent,  of  a  mill  should  sell  33^  per 
cent,  of  his  share  for  $  450,  what  would  be  the  value  of  the 
whole  mill  ? 

10.  12^  per  cent,  of  the  length  of  a  certain  railroad  is  equal 
to  3m.  lfur.  lrd.     What  is  its  entire  length  ? 

Ans.  25m.  Ofur.  8rd. 

11.  Gave  to  a  benevolent  society  19  bushels  of  corn,  which 
was  17^-  per  cent,  of  all  I  raised.  How  many  bushels  had  I 
left  ?  Ans.  9 If  bushels. 

12.  Dalton  says  to  Turner,  $36.89  is  13|  per  cent,  of  the 
sum  you  borrowed  of  me ;  and  Turner  replies,  It  is  just  16|  per 
cent,  of  the  amount  I  have  repaid  you.  How  much  of  the 
money  that  was  borrowed  remains  unpaid?        Ans.  $  57.66. 

350.  To  find  the  number  on  which  the  percentage  is  reck- 
oned, when  the  amount,  or  the  remainder,  and  the  rate  per  cent, 
are  given. 

Ex.  1.  Sold  a  horse  for  $  200,  which  was  25  per  cent,  more 
than  he  cost.     What  did  he  cost  ?  Ans.  $  160. 

operation.  Since    the 

1  _|_  .25  =  1.25  ;  $  200  —  1.25  =  $  160,  Ans.  $  200  is  ev- 
idently the 
cost  and  25  per  cent,  of  the  cost,  it  must  equal  the  cost  taken  125 
times.  Therefore,  the  given  amount,  $  200,  divided  by  1  increased 
by  .25,  or  the  given  per  cent,  expressed  decimally,  equals  $  160,  the 
basis  of  percentage  required. 


256  PERCENTAGE. 

2.  Sold  160  cords  of  wood,  which  was  20  per  cent,  less  than 
the  whole  number  owned.     How  many  cords  were  owned  ? 

Ans.  200  cords. 

operation.  Since      the 

1  —  .20  ==  .80;  160  -f-  .80  =  200  cords,  Ans.    160  cords  are 

20  per  cent, 
less  than  the  whole  number  owned,  it  must  equal  the  whole '  number 
of  cords  taken  .80  times.  Therefore  the  given  remainder,  160  cords, 
divided  by  1  decreased  by  .20,  or  the  given  per  cent,  expressed  deci- 
mally, equals  200  cords,  the  basis  of  percentage  required. 

Rule.  —  Divide  the  given  amount  by  1  increased  by  the  given  per 
cent.,  or  the  given  remainder  by  1  decreased  by  the  given  per  cent,  ex- 
pressed decimally.     Or, 

As  1  increased  by  the  given  per  cent.,  or  1  decreased  by  the  given  per 
cent.,  expressed  decimally,  is  to  1,  so  is  the  given  amount  or  the  given  re- 
mainder to  the  basis  of  percentage  required. 

Examples. 

3.  126  is  5  per  cent,  more  than  what  number  ?  Ans.  120. 

4.  328£  is  10  per  cent,  less  than  what  number  ? 

Ans.  365. 

5.  $  19.50  is  7  per  cent,  less  than  how  many  dollars? 

6.  $  34.40  is  £  per  cent,  more  than  how  many  dollars  ? 

Ans.  $34.27^|. 

7.  A  man  expends  in  a  week  $  24,  which  exceeds  by  33  J 
per  cent,  his  earnings  in  the  same  time.  What  were  the  earn- 
ings? Ans.  $18. 

8.  D.  Chandler  lives  distant  from  the  village  2m.  6  fur.  24rd., 
which  is  12  J  per  cent,  nearer  than  is  J.  Mitchel's  residence.  At 
what  distance  does  Mitchel  live  from  the  village  ? 

Ans.  3m.  lfur.  33frd. 

9.  Bought  a  carriage  for  $  123.16,  which  was  16  per  cent, 
less  than  I  paid  for  a  horse.  How  much  was  paid  for  the 
horse  ? 

10.  A  carpet  having  been  cut  from  a  piece  of  carpeting, 
there  were  left  6yd.  lTVqr.,  which  was  75  per  cent,  less  than 
the  quantity  cut  off.  How  many  yards  were  there  in  the  piece 
at  first?  Ans.  31yd.  lqr.  2na. 

11.  A  steamer  with  a  cargo  of  flour,  having  been  lightened, 
during  a  storm,  of  10  per  cent,  of  her  freight,  at  the  end  of  the 


PERCENTAGE.  257 

voyage  had  but  279  barrels  to  deliver.     How  many  barrels 
were  taken  aboard,  and  how  many  were  lost  ? 

Ans.  Taken  310  barrels ;  lost  31  barrels. 

12.  A  and  B  each  received  the  same  sum.  A  spent  86J- 
per  cent,  of  his  money  for  land,  and  B  lost  of  it  by  gambling  as 
much  as  would  equal  27^-  per  cent,  of  what  both  received. 
They  then  together  had  left  just  $  36.85^.  What  was  the  sum 
received  by  each,  and  how  much  had  each  left  ? 

Ans.  Each  received  $  63  ;  A  had  left  $  8.50J- ;  and  B  had 
left  $  28.35. 

MISCELLANEOUS  EXAMPLES. 

1.  What  is  7  per  cent,  of  1672  ?  Ans.  117^. 

2.  What  is  5£  per  cent,  of  $  3266  ?  Ans.  174ff • 

3.  Find  312  per  cent,  of  £  of  2  J. 

4.  Find  1 J  per  cent,  of  180.  Ans.  2TV 

5.  $$  is  what  per  cent,  of  3  ?  Ans.  31 6§. 

6.  -|-  is  what  per  cent,  of  T5B  ?  Ans.  40. 

7.  Find  what  per  cent.  3.50  is  of  50. 

8.  13  is  16  per  cent,  of  what  number?  Ans.  81  J. 

9.  $  66  is  100  per  cent,  of  what  number  ?        Ans.  $  66. 

10.  ^  is  61  per  cent,  of  what  number  ?  Ans.  5f -|. 

11.  $  21.28£  is  3£  per  cent,  less  than  what  sum? 

Ans.  $  22.00. 

12.  191b.  12oz.  is  16£  per  cent,  of  how  many  pounds? 

Ans.  117|-f  pounds. 

13.  A  grocer  bought  6  boxes  of  eggs,  each  containing  30 
dozen,  and  found  that  15  per  cent,  of  the  whole  were  bad ;  how 
many  eggs  did  he  lose  ?  Ans.  324  eggs. 

14.  There  is  paid  for  sawing  a  cord  of  wood  $  0.69,  which 
is  12  per  cent,  of  the  cost  of  the  wood.  What  did  the  wood 
cost? 

15.  Three  men  agreed  to  excavate  40500  cubic  feet  of 
earth;  by  the  first  week's  labor  they  excavate  200  cubic 
yards  ;  by  the  second,  6000  cubic  feet ;  and  by  the  third,  25  per 
cent,  of  what  remained  at  the  end  of  the  second  week.  They 
then  called  the  work  half  done ;  but  how  many  cubic  feet  did 
the  job  lack  of  being  half  completed  ?     Ans.  1575  cubic  feet. 

22* 


258  PERCENTAGE. 

16.  25  per  cent,  of  £  of  a  ship  is  how  many  per  cent,  of  £ 
of  it  ?  Ans.  16§  per  cent. 

17.  If  molasses  cost  20  per  cent,  less  than  $  0.50  per  gallon, 
and  it  be  sold  at  25  per  cent,  more  per  gallon  than  it  cost,  at 
what  price  is  it  sold  ?  Ans.  $  0.50  per  gal. 

18.  I  have  20  yards  of  yard-wide  cloth,  which  will  shrink  on 
sponging  4  per  cent,  in  the  length,  and  5  per  cent,  in  the 
width ;  how  much  less  than  20  square  yards  will  there  be  of  it 
after  sponging  ?  Ans.  l^f  yards. 

19.  A  gentleman  having  a  large  farm  gave  15  per  cent,  of  it 
to  his  oldest  daughter,  10  per  cent,  of  what  remained  and  ^  of 
an  acre  he  gave  to  his  oldest  son,  and  25  per  cent,  of  the  re- 
mainder he  gave  to  his  wife.  The  residue  he  divided  equally 
among  his  other  5  children,  who  received  each  39  acres.  How 
many  acres  did  his  farm  contain  ?  Ans.  340  acres. 

20.  If  the  population  of  the  United  States  in  1858  be 
30,500,000,  what  will  it  be  in  1868,  allowing  the  increase 
should  be  at  the  rate  of  34£  per  cent.  ?  t  ° 

21.  In  a  certain  battle  in  which  the  English,  French,  and  v' 
Turks  were  allied  against  the  Russians,  there  were  33£  per 
cent,  more  French  than  English,  and  the  Turks  were  8£  per 
cent,  more  than  the  French  and  1600  more  than  the  English, 
llcquired  the  whole  number  of  the  allies,  and  the  per  cent,  the 
English,  the  French  and  the  Turks  each  were  of  that  number. 

Ans.  Whole  number  13600 ;  English,  26T87  per  cent. ;  French, 
35 T^  per  cent. ;  and  Turks,  38^T  per  cent. 

22.  Bought  a  cargo  of  flour,  consisting  of  560  barrels,  at 
$7.25  per  barrel,  less  10  per  cent.,  and  sold  the  same  at  10  per 
cent,  more  than  $  7.25  per  barrel.  At  what  per  cent,  above 
the  cost  was  the  flour  sold?  How  much  was  made  by  the 
operation  ? 

23.  The  population  of  a  certain  city,  whose  gain  of  inhab- 
itants in  5  years  has  been  25  per  cent.,  is  87500 ;  what  was  it 
5  years  ago  ?  Ans.  70000. 

24.  Bought  a  horse,  buggy,  and  harness  for  $  500.  The 
horse  cost  37  J  per  cent,  less  than  the  buggy,  and  the  harness 
cost  70  per  cent,  less  than  the  horse.  "What  was  the  price  of 
each  ? 

Ans.  Buggy  $  275|f ;  horse,  $  172Jf  ;  and  harness,  $  51f  J. 


INTEREST.  259 


INTEREST. 


351.  Interest  is  the  compensation  which  the  borrower  of 
money  makes  to  the  lender ;  and  it  is  generally  reckoned  as  a 
certain  rate  per  cent,  for  any  given  time,  but  usually  for  one 
year. 

The  principal  is  the  sum  lent,  on  which  interest  is  computed. 

The  amount  is  the  interest  and  principal  added  together. 

Simple  interest  is  that  reckoned  on  the  principal  only ;  and  is 
that  meant  when  the  term  interest  is  used  alone. 

Legal  interest  is  the  rate  per  cent,  established  by  law. 

Usury  is  a  higher  rate  per  cent,  than  is  allowed  by  law. 

The  legal  rate  per  cent,  varies  in  the  different  States  and  in 
different  countries. 

In  Maine,  New  Hampshire,  Vermont,  Massachusetts,  Rhode 
Island,  Connecticut,  New  Jersey,  Pennsylvania,  Delaware, 
Maryland,  Virginia,  North  Carolina,  Tennessee,  Kentucky, 
Ohio,  Indiana,  Illinois,  Iowa,.  Nebraska,  Missouri,  Kansas,  Ar- 
kansas, Mississippi,  Florida,  District  of  Columbia,  and  on  debts 
or  judgments  in  favor  of  the  United  States,  it  is  6  per  cent. 

In  New  York,  Michigan,  Wisconsin,  Minnesota,  Georgia,  and 
South  Carolina,  it  is  7  per  cent. 

In  Alabama  and  Texas,  it  is  8  per  cent. 

In  California,  it  is  10  per  cent. 

In  Louisiana,  it  is  5  per  cent. 

In  Canada,  Nova  Scotia,  and  Ireland,  it  is  6  per  cent. 

In  England  and  France,  it  is  5  per  cent. 

Note.  —  The  legal  rate,  as  above,  in  some  of  the  States,  is  only  that  which 
the  law  allows,  when  no  particular  rate  is  mentioned.  By  special  agreement 
between  parties,  in  Ohio,  Indiana,  Michigan,  Illinois,  Iowa,  Nebraska,  Missouri, 
Kansas,  Arkansas,  Louisiana,  and  Mississippi,  interest  can  be  taken  as  high  as 
10  per  cent. ;  in  Florida,  as  high  as  8  per  cent. ;  in  Texas  and  Wisconsin,  as 
high  as  12  per  cent. ;  and  in  California  and  Minnesota,  any  per  cent.  In  New 
Jersey,  by  a  special  law,  7  per  cent,  may  be  taken  in  Jersey  City  and  the 
township  of  Hoboken.  In  Vermont  7  per  cent,  may  be  taken  on  .railroad 
bonds.  Banks  in  Illinois  cannot  take  above  7  per  cent.,  and  in  Ohio,  not  above 
6  per  cent.  In  Mississippi,  above  6  per  cent,  can  be  taken  only  for  money 
lent. 

352 •  When  the  rate  of  interest  is  6  per  cent,  per  annum, 
the  interest  for  1  year  or  12  months  will  be  6  cents  on  every 


260  INTEREST. 

100  cents  on  which  it  is  reckoned,  or  T§7  of  the  principal. 
Hence,  for  2  months  or  £  of  a  year,  it  will  be  1  cent  on  every 
100  cents,  or  T^  of  the  principal ;  and  for  1  month  or  £  of  2 
months  it  will  be  5  mills,  or  ^g-  of  the  principal.  Now,  since 
the  interest  on  100  cents  for  1  month,  or  30  days,  is  5  mills,  for 
6  days,  or  -i-  of  30  days,  it  will  be  1  mill,  or  jjfajj  of  the  prin- 
cipal ;  and  since  the  interest  on  100  cents  for  2  months  is  1  cent, 
or  -lis  of  the  principal,  for  100  times  2  months,  or  200  months, 
or  16  years  8  months,  it  will  be  100  cents,  or  equal  the  whole 
principal ;  and  in  the  same  ratio  for  any  other  length  of 
time. 

The  interest  of  $  1,  at  6  per  cent,  and  the  ratio  of  the  inter- 
est to  the  principal,  for  200  months,  and  other  convenient  parts 
of  time,  is  shown  in  the  following 

Table. 

Interest  of  $  1 

For  200  mo.  =  16yr.  8mo.  is  $  1.00  equal  the  whole  principal. 

100  mo.  =  8yr.  4mo.  "  0.50  "      £  of  the  principal. 

66fmo.  —  5yr.  6|mo.  "  0.333J  "      |  "  " 

50  mo.  — i  4yr.  2mo.  "  0.25  "  |  "  " 

40  mo.  =  3yr.  4mo.  "  0.20  «  f  "  " 

33jmo.  =  2yr.  9£uio.  "  0.166f  "  |  "  " 

25  mo.  =  2yr.  lmo.  "  0.125  "       $-  "  " 

20  mo.  =  lyr.  8mo.  "  0.10  "  fa  "  " 

16|mo.  mm  lyr.  4§mo.  "  0.0831  "  fa  "  " 

12  mo.  =  lyr.  Omo.  "  0.06  "  ^  "  " 

10  mo.  =  fofayr.  "  0.05  "  fa  "  « 

8  mo.  =  fofayr.  «  0.04  "fa  "  " 

6fmo.  =  fofayr.  «  0.033 J  "  fa  «  « 

6  mo.  =  |ofayr.  "  0.03 


Toir 

5  mo.  =  -fofayr.    "  0.025  "  fa 

4  mo.  =  Jofayr.    u  0.02  "  fa 

2  mo.  =  -Jofayr.  "  0.01  "  ^ 
1  mo.  =  fofayr.   «  0.005  "  ^ 

6  d.     =  i  of  a  mo.  "  0.001  "  jfa^ 

5  d.  —  i  of  a  mo.  "  O.OOOf  «  ^ 
4d.     =  ftofamo."  O.OOOf  «  ^ 

3  d.  _  ^  of  a  ma  «  O.OOOf  «  ^ 
2d.  -  ^  of  a  mo.  «  0.000£  «  rfto 
1  d.     =  fa  of  a  mo.  «  0.0001  «  ^ 


INTEREST. 


261 


353 #  The  principal,  the  rate  per  cent.,  the  time,  and  the  in- 
terest have  such  a  relation  to  each  other  that,  any  three  of  these 
terms  being  given,  the  fourth  can  be  readily  found.  The  com- 
putations in  interest,  therefore,  admit  of  the  following  problems 
among  others :  —  I.  To  find  the  interest ;  II.  To  find  the  prin- 
cipal ;  III.  To  find  the  rate  per  cent. ;  IV.  To  find  the  time. 

354.  To  find  the  interest  of  any  sum  for  any  time  at  6  per 
cent. 

Ex.  1.  What  is  the  interest  of  $  2640  for  2  years  7  months 
and  26  days,  at  6  per  cent.  ?  Ans.  $  420.64. 


FIRST  OPERATION. 


Interest  of  $  1  for  2  years 
Interest  of  $  1  for  7  months 
Interest  of  $  1  for  26  days 

Int.  of  $  1  for  2yr.  7mo.  26d. 


=  $0.12 
=     0.0  3  5 

se     0.0  0  44- 


0.1  5  9  £ 


The  interest  of 
$2640  for  2yr. 
7mo.  26d.  will 
be  2640  times  as 
much  as  the  in- 
terest of  $  1  for 
the  given  time. 
The  interest  of 
$1  for  2  years 
will  be  twice  as 
much  as  for  1 
year,  equal  12cts.; 
and  since  the  in- 
terest for  2  months 
is  1  cent,  for  7 
months  it  will  be 
31  cents,  or  3 
cents  5  mills.  And  as  the  interest  for  6  days  is  1  mill,  for  26  days  it 
will  be  41  mills.  These  several  sums  added  together  give  the  interest 
of  $  1  at  6  per  cent,  for  the  given  time,  equal  $  0.1 59|- ,  which  taken 
2640  times,  by  multiplying  the  given  principal  by  it,  gives  $  420.64, 
the  interest  required. 


Principal,  $  2  6  4  0 

23760 
13200 
2640 
880 


Interest,  $4  20,6  40   Ans. 


SECOND    OPERATION. 


Principal, 

-J     of  the  prin.  = 

•^    of  the  prin.  = 

yeVij-ofthe  prin.  = 


$2640 


3  3  0         Interest  for  2yr.  lmo. 
8  8         Interest  for  6mo.  20d. 
2.6  4  Interest  for  6d. 

Ans.  $420.6  4  Interest  for  2yr.  7mo.  26d. 


The  time,  2yr.  7mo.  26d.,  is  equal  to  2yr.  lmo.  -j-  6mo.  20d.  -|- 
6d.  Now,  since  the  interest  on  any  sum,  at  6  per  cent,  in  200 
months  equals  the  principal,  for  2yr.  lmo.,  or  i  of  200  months,  it  will 
equal  i  of  the  principal.  We  therefore  take  1  of  the  principal, 
$  2640,  equal  $  330,  as  the  interest  for  2yr.  lmo.     Of  the  balance. of 


262  INTEREST. 

time,  6mo.  20d.,  or  6§mo.,  being  -^  of  200  months,  we  take  -fa  of 
the  principal,  equal  $88,  as  the  interest  for  the  6mo.  20d.;  and  the 
6d.  being  j-fa$  of  200  months,  we  take  101o6  of  the  principal,  equal 
$  2.64,  as  the  interest  for  the  6d.  We  add  together  the  interest  for 
the  parts  of  the  whole  time,  and  obtain,  as  by  the  first  operation, 
$  420.64  as  the  whole  interest. 

Rule  1 .  —  Find  the  interest  of  $  1  for  the  given  time,  by  reckoning 
6  cents  for  every  year,  1  cent  for  every  two  months,  and  1  mill  for 
every  6  days  ;  then  multiply  the  given  principal  by  the  number  denoting 
that  interest,  and  the  product  will  be  the  interest  required.     Or, 

Rule  2.  —  Take  such  fractional  part  or  parts  of  the  principal  as  the 
number  expressing  the  time  is  of  200  months. 

Note  1.  —  To  find  the  amount,  add  the  principal  to  the  interest. 

Note  2.  —  In  computing  interest  for  a  fractional  part  of  a  month,  the  month 
is  considered  as  consisting  of  30  days.  This  has  the  sanction  of  general  usage 
and  the  decisions  of  the  courts,  though  not  entirely  accurate. 

Note  3.  —  Questions  in  interest,  like  other  exercises  in  percentage,  may  be 
solved  by  proportion.  The  foregoing  example  admits  of  a  statement  and  solu- 
tion by  the  rule  of  compound  proportion  (Art.  340). 

Note  4.  —  It  is  customary  among  merchants  to  reject  the  mills  in  the  re- 
sults of  their  computations  of  interest,  increasing,  however,  the  number  of 
cents  by  1  when  the  decimal  of  a  cent  exceeds  5. 

Examples. 

2.  What  is  the  interest  of  $  675  for  1  year  ?  Ans.  $  40.50. 

3.  What  is  the  interest  of  $  3967.87  for  2  years  ? 

Ans.  $  476.144. 

4.  What  is  the  interest  of  $  896.28  for  3  years  ? 

Ans  S  161.33. 

5.  What  is  the  amount  of  $  716.57  for  4  years  ? 

6.  What  is  the  amount  of  $  76.47  for  7  years  ? 

Ans.  $  108.587. 

7.  What  is  the  interest  of  $  123.45  for  6  years  ? 

Ans.  $  44.442. 

8.  What  is  the  interest  of  $  750  for  12  years  ? 

Ans.  $  540. 

9.  What  is  the  interest  of  $  130  for  2  months  ? 

10.  What  is  the  interest  of  $  85  for  3  months  ? 

Ans.  $1,275. 

11.  What  is  the  interest  of  $  19.62  for  7  months  ? 

Ans.  $0.6867. 

12.  What  is  the  interest  of  $  637  for  10  months  ? 


INTEREST.  263 

13.  What  is  the  interest  of  $  1671.32  for  14  months  ? 

Ans.  $116.99. 

14.  What  is  the  interest  of  $  891.24  for  9  months  ? 

Ans.  $  40.10. 

15.  What  is  the  interest  of  $  91  for  5  days  ?  Ans.  $  0.0758. 

16.  What  is  the  interest  of  $  324.66  for  18  days  ? 

17.  What  is  the  interest  of  $  3246  for  27  days  ? 

Ans.  $  14.607. 

18.  What  is  the  interest  $  1364.24  for  1  day  ? 

19.  What  is  the  interest  of  $  6444  for  29  days  ? 

20.  What  is  the  amount  of  $  18.60  for  24  days  ? 

21.  What  is  the  interest  of  $  386.19  for  100  months  ? 

Ans.  $  193.09. 

22.  What  is  the  interest  of  $  0.75  for  75  years  ? 

Ans.  $  3.37^ 

23.  What  is  the  interest  of  $396.15  for  1  year  1  month 
and  9  days  ?  Ans.  $  26.343. 

24.  What  is  the  interest  of  $36.18  for  3  months  and  7 
days  ?  Ans.  $  0.584. 

25.  What  is  the  interest  of  $97.15  for  2  years  11  months 
and  27  days  ? 

26.  What  is  the  interest  of  $  76.89J  from  January  11,  1852, 
to  July  27,  1863  ?  Ans.  $  53.262. 

27.  What  is  the  interest  of  $98.25  from  July  4,  1856,  to 
October  19,  1859  ?  Ans.  $  19.404. 

28.  What  is  the  interest  of  $22,763  from  February  19, 
1836,  to  July  18,  1860  ? 

29.  What  is  the  interest  of  $  175.07  from  January  7,  1855, 
to  October  12,  1859  ?  Ans.  $  50.04. 

30.  What  is  the  interest  of  $  197.28J-  from  December  6, 
1852,  to  January  11,  1854  ?  Ans.  $  12.987. 

31.  What  is  the  amount  of  $  4377.15  for  3  years  ? 

32.  What  is  the  interest  of  $444.60  for  5  years  and  6 
months  ?  Ans.  $  146.718. 

355  •     To  find  the  interest  of  any  sum  of  money  at  any  rate 
per  cent,  for  any  given  time. 

Ex.  1.     What  is  the  interest  of  $  84.50  at  7  per  cent,  for  2 
years  5  months  and  12  days?  Ans.  $  14.49. 


264  INTEREST. 

first  operation.  Having  found  the 

Principal,  $  8  4.5  0  interest  for  1  year, 

Rate  per  cent.  .0  7  and  then  for  2  years, 

the  interest  for  5 
Interest  for  1  year,  5.9  1  5  0  months  is  obtained 

2  by  first  taking  J  of 

Int.  for  2  years,  11.8300  j    ^   j"461"08*' 

t  j.  r      a  %  .* i  ,  Aff1/,  for  4  months,  and 

Int.  for  4mo.,  or  i  of  lyr.    1.9  7  1  6  +  thcn  i  of  this  last 

Int.  for  lmo.,  or  £  of  4mo.     .4929+  interestforl  month. 

Int.  for  10d.,  or  £  of  lmo.       .1  6  4  3+  And  since  10  days 

Int.  for  2d.,  or  £  of  lOd.         .0328+  are  £  of  1  month, 

Int.for2yr.5mo.12d.   $  1  4.4  9  1  6+ Ans.  ^ontSLelestVr 

the  interest  of  10 
days ;  and  since  2  days  are  £  of  10  days,  we  take  \  of  the  last  interest 
for  2  days.  The  interest  as  found  for  the  several  parts  of  the  whole 
time,  added  together,  gives  the  interest  required. 

second  operation.  We  first  find  the  in- 
Principal,  $  8  4.5  0  terest  on  the  given  sum 
Int.  of  $  1  at  6  per  cent.          .14  7  at  6  per  cent.,  and  then 

^  add  to  this  interest  the 

0  J  1  0  0  fractional  part  of  itself, 

3  3  8  0  0  denoted  by  the  excess 

8  4  5  0  of  the  rate  above  6  per 

T  ,      ,  n  ■%  c%  a  c%  i  k.  t\  ccnt-     This  excess  is  1 

Int.  at  6  per  cent        1  2.4  2  1  5  0  cent   therefore  we 

i  of  int.  at  6  per  cent.     2.0  7  0  2  5  a(M  j  0f  the  interest  at 

Int.  at  7  per  cent.     $  1  4.4  9  1  7  5  Ans.    ?  P^  ^ent- to  that  in" 
r  terest  for  the  answer. 

If  the  rate  per  cent,  had  been  less  than  6  per  cent,  we  should 

have  subtracted  the  fractional  part. 

Rule  1.  —  First  find  the  interest  for  one  year  by  multiplying  tlie  prin- 
cipal by  the  rate  per  cent,  expressed  decimally ;  and  for  two  or  more 
years  multiply  this  product  by  the  number  of  years. 

Find  the  interest  for  months  by  taking  the  most  convenient  fractional 
part  or  parts  o/one  year's  interest. 

Find  the  interest  for  days  by  talcing  the  most  convenient  fractional 
part  or  parts  o/one  month's  interest.     Or, 

Rule  2.  —  Find  the  interest  of  the  given  sum  at  6  per  cent.,  and  then 
add  to  this  interest,  or  subtract  from  it,  such  a  fractional  part  of  itself 
as  the  given  rate  is  greater  or  less  than  6  per  cent. 

Note  1.  —  I  of  the  interest  at  6  per  cent,  may  be  taken  for  that  at  1  per 
cent. ;  ^,  for  that  at  1^  per  cent. ;  £,  for  that  at  2  per  cent. ;  £,  for  that  at  3  per 
cent.  From  the  interest  at  6  per  cent,  may  be  taken  £  of  itself  for  that  at  4 
per  cent. ;  £  of  itself  for  that  at  4£  per  cent. ;  and  £  of  itself  for  that  at  5  per 
cent.    To  the  interest  at  6  per  cent,  may  be  added  I  of  itself  for  that  at  7  per 


INTEREST.  265 

cent. ;  £  of  itself  for  that  at  7£  per  cent. ;  &  of  itself  for  that  at  8  per  cent  ;  and 
£  of  itself  for  that  at  9  per  cent.  If  the  rate  is  12  per  cent.,  the  interest  at  6 
per  cent,  may  be  taken  twice;  if  18  per  cent.,  the  interest  at  6  per  cent,  may 
be  taken  three  times,  etc. 

Note  2.  —  When  in  this  book  tho  rate  of  interest  is  not  given,  6  per  cent,  is 
to  be  understood. 

Examples. 

2.  What  is  the  interest  of  $  16.75  for  7  months  and  17  days, 
at  7  per  cent.  ?  Ans.  $  0.739. 

3.  What  is  the  interest  of  $  11.10$  from  April  17,  1852,  to 
December  7,  1852,  at  7  per  cent.  ?  Ans.  $  0.496. 

4.  What  is  the  interest  of  $  12.69,  from  January  2,  1853,  to 
August  30,  1854,  at  7  per  cent.  ?  Ans.  $  1.47 J. 

5.  What  is  the  interest  of  $5000  for  2  years  5  months  26 
days,  at  7  per  cent.  ? 

6.  What  is  the  amount  of  $  416  for  3  years  16  days,  at  7  per 
cent.  ?  Ans.  $  504.64. 

7.  What  is  the  interest  of  $336  for   15    days,  at  5  per 
cent.  ?  Ans.  $  0.70. 

8.  What  is  the  interest  of  $17869.75  from  February  7, 
1852,  to  January  11,  1860,  at  5  per  cent.  ? 

Ans.  $7083.3703. 

9.  What  is  the  interest  of  $  300.50  for  1  year  2  months  and 
15  days  ?  Ans.  $  21.786. 

10.  What      the  interest  of  $  37  for  29  days,  at  5  per  cent.  ? 

11.  What  is  the  interest  of  $  35.61  from  November  11, 1861, 
to  December  15,  1863  ?  Ans.  $  4.474. 

12.  What  is  the  interest  of  $  16.76  from  December  17, 1841, 
to  January  17,  1852  ?  Ans.  $  10.558. 

13.  What  is  the  interest  of  $  1728.19  from  May  7,  1854,  to 
July  17,  1860,  at  I  per  cent.  ?  Ans.  $  26.762. 

14.  What  is  the  interest  of  $  397.16  for  1  year  6  months  and 
1  day,  at  5£  per  cent.  ?  Ans.  $  32.826. 

15.  What  is  the  amount  of  $  100.25  for  2  months  and  29 
days,  at  4  per  cent.  ?  Ans.  $  101.241. 

16.  What  is  the  interest  of  $51.17  for  9  months  and  29 
days,  at  4  per  cent.  ?  Ans.  $  1.699. 

17.  What  is  the  interest  of  $42.20  for  1  year  and  16  days, 
at  4£  per  cent.  ? 

N     23 


266  INTEREST. 

18.  What  is  the  interest  of  $  16.25  for  2  years,  at  3  per 
cent.  ?  Ans.  $  0.975. 

19.  What  is  the  interest  of  $  96.84  from  November  27, 
1849,  to  July  3,  1852,  at  7£  per  cent.  ?  Ans.  $  18.883. 

20.  What  is  the  interest  of  $  786.97  from  October  19, 1857, 
to  August  17,  1861,  at  7£  per  cent.? 

21.  What  is  the  interest  of  $  71.091  from  July  29,  1853,  to 
June  19,  1857,  at  12  per  cent.  ?  Ans.  $  33.175. 

22.  What  is  the  amount  of  $  369.29  for  2  years  3  months 
and  1  day,  at  9  per  cent.  ?  Ans.  $  444.163. 

23.  What  is  the  interest  of  $  76.35  for  1  year  8  months  and 
18  days?  Ans.  $7,864. 

24.  What  is  the  interest  of  $  47.15  for  1  month  and  19  days, 
at  13£  per  cent.?  Ans.  $  0.886. 

25.  What  is  the  interest  of  $36.72  from  May  16,  1829,  to 
February  18,  1857,  at  7  per  cent.  ?  Ans.  $  71.342. 

26.  What  is  the  interest  of  $  35.50  for  3  years  5  months  and 
20  days,  at  7  per  cent.  ?  Ans.  $  8.628. 

27.  What  is  the  amount  of  $  496.30  for  6  months  and  20 
days,  at  7  per  cent.  ?  Ans.  $  515.60. 

28.  What  is  the  interest  of  $  691.04  for  1  month  3  days,  at 
5  per  cent.  ?  Ans.  $3,167. 

29.  What  is  the  interest  of  $  9750  for  4  months,  at  2  per 
cent,  a  month  ?  Ans.  $  780. 

30.  What  is  the  interest  of  $  9162  for  3  months,  at  1£  per 
cent,  a  month  ?  Ans.  $  412.29. 

31.  What  is  the  interest  of  $  1500  for  7  months  20  days,  at 
10  per  cent.  ?  Ans.  $  95.833. 

32.  What  is  the  interest  of  $  640.50  for  10  months  and  26 
days,  at  10  per  cent.  ?  Ans.  $  58.00. 

33.  What  is  the  interest  of  $3178  for  15  months  and  15 
days  ? 

34.  If  a  banker  borrow  $  10,000  in  Boston  at  6  per  cent., 
and  let  the  same  in  Wisconsin  at  7  per  cent.,  how  much  does 
he  make  by  the  operation  in  that  way  in  2  years  and  6f 
months  ?  Ans.  $  255.555. 

356.  To  reckon  interest  for  any  number  of  days,  when  12 
months  of  only  30  days  each,  or  360  days,  are  considered  a 
year. 


INTEREST. 


267 


Ex.  1.     What  is  the  interest  of  $  460  for  93  days  ? 

Ans.  $7.13. 


FIRST    OPERATION. 


Principal, 

i  of  93d.  =  I5i&. 


We  multiply  the  principal 
$  4  6  0  by  i  of  the  number  of  days, 

.0  1  5  J-  considered  as  thousandths; 

since  the  interest  of  $  1  at  6 
per  cent,  for  6  days  is  1  mill 
or  1  thousandth  of  a  dollar, 
(Art.    352,)    and    for  any 


2300 
460 
230 

Interest,  $  7.1  3  0  Ans.    other  number  of  days,  at 

the  same  per  cent,  must  be 
one  sixth  as  many  mills  or  thousandths  of  a  dollar  as  there  are 
days 


SECOND    OPERATION. 


Principal, 

y^g-  of  the  principal, 


J    of  the  interest  for  60d. 
of  the  interest  for  30d. 


tV 


$460 

4.6  0 

2.3  0 

.2  3 


Ans.   $7.13 


Interest  for  60d. 
Interest  for  30d. 
Interest  for    3d. 

Interest  for  93d. 


93  days  =60  days  -\-  30  days  -f-  3  days.  Now,  since  the  interest 
of  any  sum  at  6  per  cent,  for  2  months,  or  60  days,  equals  yj-g-  of  the 
principal  (Art.  352),  ^  of  $  460  =  $4.60  will  be  the  interest  for 
60  days  ;  \  of  the  interest  for  60  days,  or  $  2.30,  will  be  that  for  30 
days;  and  J*  of  the  interest  for  30  days,  or  $  0.23,  will  be  that  for  3 
days.  The  interest  for  the  several  parts  of  the  time  added  together 
must  give  the  interest  for  the  whole  time,  or  93  days. 

Rule.  —  Multiply  the  principal  by  one  sixth  of  the  number  of  days 
expressed  decimally  as  thousandths.    Or, 

Divide  the  principal  by  100,  and  take  such  a  part  or  parts  of  the  quo- 
tient as  the  given  number  of  days  is  of  60  days. 

The  result  will  be  the  interest  at  6  per  cent.,  from  which  it  may  be 
found  for  any  other  rate,  as  in  Art.  355. 

Note  1.  —  One  sixth  of  the  number  of  days  in  any  number  of  months  of  30 
days  each  is  equal  to  5  times  the  number  of  months.  Thus,  one  sixth  of  the 
number  of  days  in  7  months  equals  7  X  5,  or  35  days ;  and  one  sixth  of  the 
number  of  days  in  9  months  and  13  days  equals  (9X5)  +  (13  -r  6),  or  45  + 
2*  =  471  days. 

Note  2.  —  It  is  a  common  practice  among  mercantile  men,  in  calculating 
interest,  to  consider  a  year  to  consist  of  only  360  days,  but  the  laws  of  some  of 
the  States  require  365  days  to  a  year.  360  days  are  in  fact  only  2|°  =  l|  0f  a 
common  year,  and  therefore  are  Jg  less  than  what  perfect  accuracy  would  re- 
quire. Hence,  when  the  year  is  considered  to  be  one  of  365  days,  the  interest  as 
found  by  the  rule  must  be  diminished  by  -%  of  itself ;  Or,  we  may  multiply  the 
principal  by  the  number  of  days,  and,  if  the  rate  be  6  per  cent.,  divide  by  6083  J; 
if  the  rate  be  7  per  cent.,  divide  by  5214;  and  if  the  rate  be  5  per  cent.,  di- 
vide by  7300. 


268  INTEREST. 

Examples. 

2.  What  is   the   interest   of  $96   for  33   days,  at  7  per 
cent.?  Ans.  $.616. 

3.  What  is  the  interest  of  $  320.40  for  63  days,  at  5  per 
cent.?  Ans.  $2,803. 

4.  What  is  the  interest  of  $  131.20  for  123  days  ? 

Ans.  $2,689. 

5.  What  is  the  interest  of  $  26.60  for  78  days? 

6.  What  is  the  interest  of  $5780  for  153  days,  at  10  per 
cent?  Ans.  $245.65. 

7.  What  is  the  interest  of  $  105.10  for  48  days,  at  12  per 
cent.?  Ans.  $1,681. 

8.  What  is  the  interest  of  $  13.62  for  93  days,  at  7£  per 
cent.  ? 

9.  What  is  the  interest  of  $  4580  for  253  days,  at  5£  per 
cent.?  Ans.  $  177.029f|. 

10.  What  is  the  interest  of  $3140  for  273  days,  at  7  per 
cent.?  Ans.  $  166.681§. 

11.  What  is  the  interest  of  $  10550  for  243  days,  at  8  per 
cent.?  Ans.  $569.70. 

12.  What  is  the  amount  of  $  33.44  for  333  days  ? 

Ans.  $  35.295. 

13.  What  is  the  amount  of  $71.60  for  3  months  and  18 
days  ? 

14.  What  is  the  interest  of  $  92.96  for  4  months  and  3  days, 
at  7  per  cent.  ?  Ans.  $  2.223. 

15.  What  is  the  interest  of  $  144.50  for  144  days,  allowing 
365  days  to  the  year,  at  5  per  cent.  ?  Ans.  $  2.85. 

16.  What  is  the  interest  of  $  761.81  for  165  days,  allowing 
365  days  to  the  year?  Ans.  $ 20.662. 

17.  What  is  the  interest  of  $560  for  183  days,  allowing  365 
days  to  a  year,  at  7  per  cent.  ?  Ans.  $  19.65. 

18.  What  is  the  interest  of  $  1960  for  93  days  ? 

19.  What   is   the   amount   of  $1000  for  1  month  and  3 
days?  Ans.  $1005.50. 

20.  What  is  the  amount  of  $7300  for  18  days,  at  6  per 
cent.?  Ans.  $7321.90. 


INTEREST.  269 

357.  To  find  the  interest  on  sterling  money,  at  any  rate  per 
cent.,  for  any  time. 

Ex.  1.  What  is  the  interest  of  576£.  5s.  6d.  for  1  year  and 
10  months  ?  Ans.  63£.  7s.  9d. 

OPERATION. 

576£.  5s.  8£d.  =  5  7  6.2  5  5  £.    Principal. 

Tiy  of  the  principal,  5  7.6  2  5  5      Int.  for  lyr.  8mo. 

T\y  of  the  int.  for  lyr.  8mo.       5.7  6  2  5  5  Int.  for  2mo. 

Ans.  6  3.3  8  8  0  5  £.  = 

63£.  7s.  9d.-f-  Int.  for  lyr.lOmo. 

We  reduce  the  5s.  6d.  to  the  decimal  of  a  pound  (Art.  279),  and 
annex  it  to  the  pounds ;  we  then  find  the  interest  as  though  the  sum 
expressed  dollars  and  cents ;  and,  reducing  the  decimal  in  the  answer 
to  shillings  and  pence,  we  have  63£.  7s.  9d.-{-  as  the  interest  re- 
quired.    Hence, 

Reduce  the  shillings,  pence,  and  farthings,  if  any,  in  the  principal  to 
the  decimal  of  a  pound.  Then  proceed  as  in  United  States  money; 
and,  if  there  be  a  decimal  in  the  result,  reduce  it  to  a  compound  number. 

Examples. 

2.  What  is  the  interest  of  179£.  12s.  lid.  for  1  year  and  7 
months,  at  5  per  cent.?  Ans.  14£.  4s.  5|-d. 

3.  What  is  the  interest  of  25£.  for  1  year  and  9  months,  at 
5  per  cent.  ? 

4.  What  is  the  interest  of  5440£.  10s.  for  3  years  and  11 
months,  at  6  per  cent.  ?  .   Ans.  1278£.  10s.  4d. 

5.  What  is  the  interest  of  943£.  Is.  8d.  from  May  1, 1857,  to 
October  21,  1857,  allowing  365  days  to  a  year,  at  5£  per 
cent.  ?  Ans.  23£.  9s.  4d. 

358.  To  find  the  principal,  the  interest,  the  time,  and  the 
rate  per  cent,  being  given. 

Ex.  1.  What  principal  will  gain  $  120  in  4  years,  at  6  per 
cent.  ?  Ans.  $  500. 

operation.  The  interest  for  the  given 

$12  0.0  0~-.2  4  =  $500  Aois.     time  and  rate  is  24  cents  on 

a  principal  of  $  1  ;  therefore, 
the  required  principal  will  be  as  many  times  $  1  as  the  number  denot- 
ing the  given  interest  contains  times  .24. 

Rule.  —  Divide  the  given  interest  by  the  interest  of  $  1  for  the  given 
time,  at  the  given  rate. 

23* 


270  INTEREST. 

Examples. 

2.  What  principal  will  gain   $  10.08  in  1  year,  at  7  per 
cent.?  Ans.  $144. 

3.  What  principal  will  yield  an  income  of  $  13.20  in  1  year 
and  4  months,  at  8£  per  cent.  ?  Ans.  $  120. 

4.  What   sum   should  be  paid   for  a  ground   rent   paying 
yearly,  at  6  per  cent.,  $  40.50  ? 

5.  If  the  interest  on  a  sum  borrowed  at  2  per  cent,  a  month 
is  $  24  for  90  days,  what  is  the  sum  ?  Ans.  $  400. 

6.  What  principal  at  7£  per  cent,  is  sufficient  to  produce 
$  206.38£  interest  in  183  days  ?  Ans.  $  5600. 

359.     To  find  the  rate  per  cent.,  the  principal,  the  in- 
terest, and  the  time  being  given. 

Ex.  1.     If  the  interest  of  $500  for  4  years  is  $120,  what 
is  the  rate  per  cent.  ? 

operation.  "We   fmrt  the  interest  on   a 

120  -r  20=  G  per  cent.  Ans.     given  principal  for  the  given 

term  to  be  $  20,  at  1  per  cent. ; 
then  the  required  rate  will  be  as  many  per  cent,  as  20  is  contained 
times  in  120,  or  6  per  cent. 

Rule.  —  Divide  the  given  interest  by  the  interest  of  the  given  princi- 
pal at  1  per  cent,  for  the  given  time. 

Examples. 

2.  The  interest  of  $  144  for  1  year  is  $  10.08  ;  what  is  the 
rate  per  cent.  ?  Ans.  7  per  cent. 

3.  At  what  rate  per  cent,  must  $  120  be   on   interest   to 
amount  to  $  133.20  in  1  year  4  months? 

4.  At  what  rate  per  cent,  must  $  l,or  any  other  sum,  be  on 
interest  to  double  itself  in  14f  years?  Ans.  7  per  cent. 

5.  At  what  rate  per  cent,  must  any  sum  of  money  be  on  in- 
terest to  quadruple  itself  in  33£  years  ?  Ans.  9  per  cent. 

6.  I  receive  yearly  $  232.50  interest  on  $  4650  loaned  to 
the  State  of  Massachusetts ;  what  is  the  rate  per  cent.  ? 

Ans.  5  per  cent. 

7.  At  what  rate  per  cent,  must  $7500  be  loaned  to  gain 
$  60.00  in  48  days  ? 


INTEREST.  271 

8.  At  what  rate  per  cent,  must  $  280  be  on  interest  to  amount 
to  $  411.95  in  6  years  and  6  months  ?  Ans.  7  £  per  cent. 

9.  At  what  rate  per  cent,  must  $  480  be  on  interest  to  amount 
to  $529.60  in  1  year  3  months  and  15  days? 

Ans.  8  per  cent. 

300.  To  find  the  time,  the  principal,  the  interest,  and  the 
rate  per  cent,  being  given. 

Ex.  1.     How  long  must  $  500  be  on  interest,  at  6  per  cent., 

to  gain  $  120  ?  Ans.  4  years. 

operation.  We  find  that  the  given  princi- 

120  —  30  =  4  years,  Ans.     pal  at  the  given  rate  will  produce 

$  30  interest  in  1  year ;  therefore 
to  produce  $  1 20  interest  there  will  be  required  as  many  years  as 
30  is  contained  times  in  120,  or  4  years. 

Rule.  —  Divide  the  given  interest  by  the  interest  of  the  given  princi- 
pal for  1  year. 

Examples. 

2.  How  long  must  $  120  be  on  interest,  at  8  j-  per  cent.,  to 
gain  $  13.20  ?  Ans.  lyr.  4mo. 

3.  In  what  time  will  $  144  produce  $  10.08  interest,  at  7  per 
cent.  ?  Ans.  1  year. 

4.  In  what  time  will  $  240  at  interest  amount  to  $  280,  at  6 
per  cent.  ?  Ans.  2yr.  9mo.  lOd. 

5.  In  what  time  will  $  1,  or  any  other  sum,  double  itself,  at  5 
per  cent,  interest  ?  Ans.  20  years. 

6.  In  what  time  wdll  any  sum  double  itself  at  10  per  cent,  in- 
terest? Ans.  10  years. 

7.  In  what  time  will  $  1500  amount  to  $  2250,  at  5  per  cent, 
interest  ? 

8.  In  what  time  will  $480  at  4^-  per  cent,  amount  to 
$  561.60  ?  •  Ans.  3yr.  9mo.  lOd. 

9.  In  what  time,  at  12  per  cent.,  will  $1728  amount  to 
$3853.44?  Ans.  lOyr.  3mo. 

10.  In  what  time,  at  6  per  cent.,  will  $  240  amount  to 
$  720  ?  Ans.  33yr.  4mo. 

11.  Borrowed,  May  16,  1857,  the  sum  of  $400,  payable  as 
soon  as  the  principal,  increased  by  the  interest  at  6  per  cent., 
shall  equal  $  500.     At  what  date  is  it  payable  ? 

Ans.  July  16,  1861. 


272  PROMISSORY   NOTES. 

I 

PROMISSORY  NOTES. 

361.  A  Promissory  Note,  or  note  of  hand,  is  an  engage- 
ment, in  writing,  to  pay  a  specified  sum,  either  to  a  person 
named  in  the  note,  or  to  his  order,  or  to  the  bearer. 

A  joint  note  is  one  signed  by  two  or  more  persons,  who  to- 
gether are  holden  for  its  payment. 

A  joint  and  several  note  is  one  signed  by  two  or  more  per- 
sons, who  separately  and  together  are  holden  for  its  payment. 

A  negotiable  note  is  one  so  made  that  it  can  be  sold  or  trans- 
ferred from  one  person  to  another. 

362.  The  maker  or  drawer  of  a  note  is  the  person  who 
signs  it. 

The  payee,  promisee,  or  holder  of  a  note  is  the  person  to 
whom  it  is  to  be  paid. 

The  indorser  of  a  note  is  the  person  who  writes  his  name 
upon  its  back  to  transfer  it,  or  as  a  guaranty  of  its  payment.  If 
the  indorser,  however,  wishes  only  to  transfer  the  note,  he  may 
write  before  his  name  the  words  "  without  recourse,"  and  then, 
though  by  his  name  he  guarantees  the  genuineness  of  the  note, 
he  is  not  liable  for  its  payment,  should  the  maker  not  pay  it 
when  due. 

363.  The  face  of  a  note  is  the  sum  for  which  it  is  given. 

364.  A  note  should  contain  the  words  "value  received," 
and  the  sum  for  which  it  is  given  should  be  expressed  in 
written  words,  or,  as  is  the  general  custom,  the  dollars  may  be 
written  in  words  and  the  cents,  if  any,  be  expressed  as  hun- 
dredths of  a  dollar  in  the  form  of  a  common  fraction. 

Note.  —  The  laws  of  Pennsylvania  require  that  the  words  "  without  defal- 
cation "  should  be  inserted  in  a  promissory  note;  and  in  Indiana,  notes  generally 
contain  the  words  "  without  any  relief  whatever  from  valuation  or  appraise- 
ment laws." 

365.  A  note  is  said  to  be  one  on  time,  when  it  is  made  pay- 
able on  or  after  a  certain  date,  and  the  day  on  which  it  becomes 
legally  due  is  called  the  day  of  maturity. 

366.  According  to  the  laws  of  many  of  the  States,  when  a 
particular  day  is  specified  in  the  note  for  its  payment,  three 


PROMISSORY   NOTES.  273 

days  additional  are  allowed,  called  days  of  grace,  within  which 
the  maker  may  pay  the  note,  unless  it  states  "  without  grace." 
Should  the  third  day  of  grace,  however,  fall  on  Sunday,  or  some 
public  day,  the  day  of  maturity  will  be  a  day  earlier. 

367.  When  a  note  is  written  for  months,  calendar  months 
are  understood.  Thus,  if  a  note  be  dated  April  21,  for  one 
month,  it  will  be  nominally  due  May  21 ;  and  if  dated  January 
29,  30,  or  31,  it  being  for  one  month,  it  will  be  nominally  due 
February  28,  should  it  be  a  common  year,  or-February  29, 
should  it  be  a  leap  year. 

868.  When  a  note  is  written  without  interest,  it  can  only 
draw  interest,  if  on  time,  after  the  time  specified  for  its  payment, 
and  not  then  lawfully,  in  some  States,  till  after  a  demand  has 
been  made ;  or  if  not  on  time,  after  payment  has  been  de- 
manded. 

Note.  —  A  note  attested  or  witnessed,  in  Massachusetts  and  some  other 
States,  is  taken  out  of  the  statute  of  limitation. 


PARTIAL  PAYMENTS  ON  NOTES  AND  BONDS. 

369.  Partial  or  part  payments  on  notes,  bonds,  or  other 
obligations,  being  receipted  for  by  entry  on  the  back  of  the 
obligation,  are  termed  indorsements. 

United  States  Rule. 

370.  In  the  United  States  courts,  and  the  courts  of  Mas- 
sachusetts, New  York,  and  several  other  States,  interest  on 
notes  and  bonds,  when  partial  payments  have  been  made,  is 
reckoned  according  to  the  following 

Rule.  —  Compute  the  interest  on  the  principal  to  the  time  when  the 
first  payment  was  made,  which  equals  or  exceeds,  either  alone  or  with 
preceding  payments,  the  interest  then  due. 

Add  that  interest  to  the  principal,  and  from  the  amount  subtract  the 
payment  or  payments  thus  far  made. 

The  remainder  will  form  a  new  principal ;  on  which  compute  the  in- 
terest, proceeding  as  before. 

Note.  —  This  rule  is  on  the  principle  that  neither  interest  nor  payment 
should  draw  interest. 


274  PROMISSORY  NOTES. 


Examples. 

(1.)     $  165.18.  Boston,  June  17,  1847. 

For  value  received,  I  promise  to  pay  Nathaniel  Ford,  or 
order,  on  demand,  one  hundred  and  sixty-five  dollars  and 
eighteen  cents,  with  interest. 

Attest,  Joseph  Field.  James  Peterson. 

On  this  note  are  the  following  indorsements.  December  7,  1847,  received 
eighteen  dollars  and  thirteen  cents  of  the  within  note.  October  19,  1848,  re- 
ceived twenty-eight  dollars  and  sixteen  cents.  September  25, 1849,  received 
thirty-six  dollars  and  twelve  cents.  July  10,  1850,  received  three  dollars  and 
eighteen  cents.  June  6,  1851,  received  thirty-six  dollars  and  twenty-eight 
cents.  December  28,  1852,  received  thirty-one  dollars  and  seventeen  cents. 
May  5,  1853,  received  three  dollars  and  eighteen  cents.  September  1,  1853, 
received  twenty  five  dollars  and  eighteen  cents.  October  18, 1854,  received 
ten  dollars. 

How  much  remains  due  September  27,  1855  ? 

Ans.  $15,417. 

OPERATION. 

Principal,  carrying  interest  from  June  17,  1847,  $  165.180 

Interest  from  June  17,  1847,  to  Dec.  7,  1847,  5mo.  20d.,  4.680 

Amount,       169.860 
First  payment,  Decemlfer  7,  1847,  ....       18.130 

Balance  for  new  principal, 151.730 

Interest  from  Dec.  7,  1847,  to  Oct.  19,  1848,  lOmo.  12d.,  7.889 

Amount,       159.619 
Second  payment,  October  19,  1848,  .         .         .  28.160 

Balance  for  new  principal, 131.459 

Interest  from  Oct.  19,  1848,  to  Sept.  25, 1849,  llmo.  6d.,  7.361 

Amount,       138.820 
Third  payment,  September  25,  1849,      ....         36.120 

Balance  for  new  principal, 102.700 

Interest  from  Sept.  25,  1849,  to  June  6, 1851,  20mo.  lid.,        10.458 

Amount,       113.158 
Fourth  pay't,  July  10,  1850,  a  sum  less  than  interest,      3.18 
Fifth  pay't,  June  6, 1851,  a  sum  greater  than  interest,  36.28 

39.460 


Balance  for  new  principal,      .         .         .         .         .         .  73.698 

Interest  from  June  6,  1851,  to  Dec.  28,  1852,  18mo.  22d.,  6.903 

Amount,         80.601 


PROMISSORY  NOTES.  275 

Amount  brought  forward,     $  80.601 
Sixth  payment,  December  28,  1852,        ....         31.170 

Balance  for  new  principal, 49.431 

Interest  from  Dec.  28,  1852,  to  May  5,  1853,  4mo.  7d.,  1.046 

Amount, 
Seventh  payment,  May  5,  1853, 

Balance  for  new  principal,  .... 

Interest  from  May  5,  1853,  to  Sept,  1,  1853,  3mo.  26d., 

Amount,         48.211 
Eighth  payment,  September  1,  1853,       ....         25.180 

Balance  for  new  principal, 23.031 

Interest  from  Sept.  1,  1853,  to  Oct.  18,  1854,  13mo.  17d.,  1.562 

Amount,         24.593 
Ninth  payment, 10.000 

Balance  for  new  principal, 14.593 

Interest  from  Oct.  18,  1854,  to  Sept.  27,  1855,  llmo.  9d./  .824 

Balance  due  at  the  time  of  payment,       .         .         .         .      $15,417 

(2.)     $  769.87.  St.  Louis,  June  17, 1849. 

For  value  received,  I  promise  to  pay  L.  Swan,  or  order,  on 
demand,  seven  hundred  and  sixty-nine  dollars  and  eighty-seven 
cents,  with  interest.  Samuel  Q.  Peters. 

Attest,  Moses  Haynes. 

Payments:  March  1, 1850,  seventy-five  dollars  and  fifty  cents;  June  11, 
1851,  one  hundred  and  sixty-five  dollars;  September  15,  1851,  one  hundred 
and  sixty-one  dollars;  Jan.  21, 1852,  forty-seven  dollars  and  twenty-five  cents; 
March  5, 1853,  twelve  dollars  and  seventeen  cents ;  December  6,  1853,  ninety- 
eight  dollars;  July  7, 1854,  one  hundred  and  sixty-nine  dollars. 

What  remains  due  September  25,  1855  ?  Ans.  $  226.297. 

(3.)     $  300.  Chicago,  April  30,  1851. 

For  value  received,  I  promise  Kimball  fy  Hammond  to  pay 
them,  or  order,  on  demand,  three  hundred  dollars,  with  interest. 

Simpson  W.  Leavet. 

Payments :  June  27, 1852,  one  hundred  and  fifty  dollars ;  December  9, 1852, 
one  hundred  and  fifty  dollars. 

What  was  due  October  9,  1853  ?  Ans.  $  26.735. 

(4.)     $  54.18.  San  Francisco,  Feb.  11,  1852. 

For  value  received,  I  promise  to  pay  John  Trow,  or  order,  on 
demand,  fifty-four  dollars  and  eighteen  cents,  with  interest. 

Luke  M.  Sampson. 


276  PROMISSORY   NOTES. 

Payments:  July  11, 1853,  twelve  dollars  and  twenty-five  cents;  August  15, 
1854,  two  dollars  and  ten  cents;  July  9,  1855,  three  dollars  and  twelve  cents; 
August  21, 1855,  thirty-seven  dollars  and  eighteen  cents. 

What  was  due  December  17,  1855  ?  Ans.  $  10.222. 

(5.)     $  1728.  Philadelphia,  Jan.  7,  1851. 

For  value  received,  we  jointly  and  severally  promise  to  pay 
Jones,  Oliver,  fy  Co.,  or  order,  one  thousand  seven  hundred  and 
twenty-eight  dollars,  on  demand,  with  interest,  without  defalca- 
tion. John  Wickersham. 

Attest,  Timothy  True.  James  Thickstein. 

(6.)     $  1000.  New  York,  January  1,  1850. 

For  value  received,  I  promise  to  pay  James  Johnson,  or  order, 
on  demand,  one  thousand  dollars,  with  interest  at  seven  per  cent. 

Samuel  T.  Fortune. 

Indorsements:  September  28,  1850,  one  hundred  and  forty-four  dollars; 
March  1,  1851,  twenty  dollars;  July  17,  1851,  three  hundred  and  sixty  dol- 
lars; August  9,  1851,  one  hundred  and  ninety  dollars ;  September  25,  1852, 
one  hundred  and  seventy  dollars ;  December  11,  1853,  two  hundred  dollars; 
July  4, 1855,  seventy-five  dollars. 

What  was  due  June  1,  1857  ?  Ans.  $  7.61. 

Connecticut  Rule. 

371  •  The  Supreme  Court  of  the  State  of  Connecticut  has 
adopted  the  following 

Rule.  —  Compute  the  interest  to  (lie  time  of  the  first  payment,  if  that 
be  one  year  or  more  from  the  time  the  interest  commenced ;  add  it  to  the 
principal,  and  deduct  the  payment  from  the  sum  total.  If  there  be  after 
payments  made,  compute  the  interest  on  the  balance  due  to  the  next  pay- 
ment, and  then  deduct  the  payment  as  above  ;  and  in  like  manner  from 
one  payment  to  another,  till  all  the  payments  are  absorbed;  provided  the 
time  between  one  payment  and  another  be  one  year  or  more. 

If  any  payments  be  made  before  one  year's  interest  has  accrued,  then 
compute  the  interest  on  the  principal  sum  due  on  the  obligation  for  one 
year,  add  it  to  the  principal,  and  compute  the  interest  on  the  sum  paid 
from  the  time  it  was  paid  up  to  the  end  of  the  year;  add  it  to  the 
sum  paid,  and  deduct  that  sum  from  the  principal  and  interest  added 
as  above. 

If  any  payments  be  made  of  a  less  sum  than  the  interest  arisen  at  the 
time  of  such  payment,  no  interest  is  to  be  computed,  but  only  on  the  prin- 
cipal sum  for  any  period. 

Note.  —  If  a  year  extends  beyond  the  time  of  settlement,  find  the  amount 
of  the  remaining  principal  to  the  time  of  settlement ;  find  also  the  amount  of 
the  indorsement  or  indorsements,  if  any,  from  the  time  they  were  paid  to  the 
time  of  settlement,  and  subtract  their  sum  from  the  amount  of  the  principal. 


PROMISSORY  NOTES.  277 

Example. 

(1.)     $  900.  New  Haven,  June  1,  1858. 

For  value  received,  I  promise  to  pay  J,  Downs,  or  order, 
nine  hundred  dollars,  on  demand,  with  interest 

James  L.  Emerson. 

Indorsements:  June  16,  1859,  two  hundred  dollars;  August  1,  1860,  one 
hundred  and  sixty  dollars ;  November  16,  1860,  seventy-five  dollars;  February 
1,  1862,  two  hundred  and  twenty  dollars. 

What  was  due  August  1,  1862  ?  Ans.  $  417.822. 

OPERATION. 

Principal, $900.00 

Interest  from  June  1,  1858,  to  June  16,  1859,  \2\  months,       56.25 

956.25 
First  payment, 200.00 

756.25 
Interest  from  June  16, 1859,  to  August  1, 1860,  131  months,     51.046 

807.296 
Second  payment, 160.000 

647.296 
Interest  for  1  vear, 38.837 


686.133 
Am't  of  3d  pay't,  from  Nov  16,  1860,  to  Aug.  1, 1861, 8-^mo.,     78.187 

607.946 
Interest  from  Aug.  1,  1861,  to  Aug.  1,  1862,  12  months,  36.476 

644.422 
Am't  of  4th  pay't,  from  Feb.  1, 1862,  to  Aug.  1, 1862,  6mo.,     226.600 

Balance  due  August  1,  1862,      ....      $417,822 

Merchants'  Ktjle. 

372.  It  is  customary  with  merchants  and  others,  when  par- 
tial payments  are  made  of  notes  or  other  debts,  when  the  note  or 
debt  is  settled  within  a  year  after  becoming  due,  to  adopt  the 
following 

Rule.  — Find  the  amount  of  the  principal  from  the  time  it  became 
due  until  the  time  of  settlement.     Then  find  the  amount  of  each  indorse- 
ment from  the  time  it  teas  paid  until  settlement,  and  subtract  their  sum 
from  the  amount  of  the  principal. 
N      24 


278  PROMISSORY  NOTES. 


Examples. 


(1.)     $  1728.  Baltimore,  January  1,  1853. 

For  value  received,  I  promise  to  pay  Riggs,  Peabody,  S?  Co,, 
or  order,  on  demand,  one  thousand  seven  hundred  and  twenty- 
eight  dollars,  with  interest.  John  Paywell,  Jr. 

Indorsements:  March  1,  1853,  three  hundred  dollars;  May  16,  1853,  one 
hundred  and  fifty  dollars ;  September  1,  1853,  two  hundred  and  seventy  dol- 
lars; December  il,  1853,  one  hundred  and  thirty-five  dollars. 

What  was  due  at  the  time  of  payment,  which  was  December 
16,  1853?  Ans.  $948.03 

OPERATION. 

Principal, $1728.00 

Interest  for  11  months  and  15  days,        .         .         .         .  99.36 

$  1827.36 

First  payment, $300.00 

Interest  for  9  months  and  15  days,     .  14.25 

Second  payment,          ....  150.00 

Interest  for  7  months,        .         .         .  5.25 

Third  payment,             ....  270.00 

Interest  for  3  months  and  15  days,     .  4.72 

Fourth  payment,           ....  135.00 

Interest  for  5  days,             .         .         .  .11 


879.33 


Balance  remaining  due,  December  16,  1853,     $  948.03 

(2.)     $  700.  Montpelier,  February  4,  1854. 

For  value  received,  we  jointly  and  severally  promise  to  pay 
James  Thomas,  or  order,  on  demand,  seven  hundred  dollars, 
with  interest  Sampson  Phillips, 

Kichard  Fletcher. 

Payments:  March  18,  1854,  one  hundred  and  sixty  dollars;  June  24,  1854, 
two  hundred  dollars;  September  11,  1854,  one  hundred  and  twenty  dollars ; 
October  5,  1854,  sixty  dollars. 

What  was  due  on  this  note  Nov.  28,  1854  ?     Ans.  $  180.43. 

(3.)     $  500.  Detroit,  January  1,  1857. 

For  value  received,  three  months  after  date  1  promise  to  pay 
to  the  order  of  James  Francis  five  hundred  dollars. 

William  Amsden. 

Indorsement:  July  1, 1857,  two  hundred  dollars. 
What  was  due  April  1,  1858,  the  rate  of  interest  being  7  per 
cent.?  Ans.  $324.50. 


COMPOUND   INTEREST. 


279 


COMPOUND  INTEREST. 

373.  Compound  Interest  is  interest  on  the  original 
principal  with  its  interest  added  when  remaining  unpaid  after 
becoming  due. 

374.  When  the  interest  is  added  to  the  principal  at  the  end 
of  every  year,  and  a  new  principal  is  thus  formed  yearly,  it  is 
said  to  compound  annually ;  when  the  interest  is  added  to  the 
principal  so  as  to  form  a  new  principal  half-yearly,  it  is  said  to 
compound  semiannually. 

375.  Compound  interest  is  based  upon  the  principle,  that,  if 
the  borrower  does  not  pay  the  interest  as  it  becomes  due  at 
stated  times,  it  is  no  more  than  just  for  him  to  pay  interest  for 
the  use  of  it,  so  long  as  he  shall  have  it  in  his  possession. 

Note.  —  Compound  interest  is  not  favored  by  the  laws,  though  it  is  not 
usurious.  A  contract  or  promise  to  pay  money  with  compound  interest  can- 
not generally  be  enforced,  being  only  valid  for  the  principal  and  legal  interest. 

376.  To  find  the  compound  interest  of  any  sum  of  money 
at  any  rate  per  cent,  for  any  time. 

Ex.  1.   What  is  the  compound  interest  of  $  300  for  3  years? 


FIRST    OPERATION. 


Principal  for  1st  year, 
Interest  of  $  1  for  1  year, 

Interest  for  1st  year, 
Principal  for  2d  year, 
Interest  for  2d  year, 
Principal  for  3d  year, 
Interest  for  3d  year, 


$300 
.0  6 

18.0  0 
3  0  0.0  0 

3  1  8.0  0 
.0  6 

1  9.0  8  0  0 
318.0  0 


3  3  7.0  8 
.06 


2 
33 


0.2  2  4  8 

708 


Amount  for  3  years,      3  5 
First  principal,  3  0 

Comp.  int.  for  3  years,  $  5  7.3  0  4  8,  Ans. 


7.3  0  4  8 
0.0  0 


We  first  multiply  the 
given  principal  by  the 
number  denoting  the 
interest  of  $  1  for  one 
year,  and  add  the  in- 
terest thus  found  to 
the  principal  for  the 
amount ;  on  which  as  a 
new  principal  we  find 
the  interest  for  the  sec- 
ond year,  and  proceed  as 
before ;  and  so  also  with 
the  third  year.  From 
the  amount  of  the  last 
year  we  subtract  the 
first  principal,  and  ob- 
tain the  compound  in- 
terest for  3  years. 


280  COMPOUND   INTEREST. 


SECOND    OPERATION- 


Principal  for  1st  year,  $  3  0  0.0  0 

J%  of  the  principal,  1  5.0  0 

-£  of  the  interest  at  5  per  cent,      3.0  0 

Principal  for  2d  year,  3  1 8.0  0  Amount  for  1st  year  at 

2*y  of  the  principal,  1  5.9  0  [6  per  cent, 

i  of  the  interest  at  5  per  cent.,      3.1  8 

Principal  for  3d  year,  3  3  7.0  8  Amount  for  2d  year  at 

2-\r  of  the  principal,  1  6.8  4  5  [6  per  cent. 

■J  of  the  interest  at  5  per  cent.,      3.3  7  0 

3  5  7.3  0  4  Amount  for  3d  year 
First  principal,  3  0  0.0  0  [at  6  per  cent. 

Compound  interest  for  4  years,  $  5  7.3  0  4  Ans. 

In  the  second  operation  the  work  is  somewhat  abridged,  by  finding 
the  interest  for  each  year  at  6  per  cent,  by  taking  -^  of  the  principal 
for  the  interest  at  5  per  cent.,  and  \  of  that  for  interest  at  1  per 
cent. 

Rule.  —  Find  the  interest  of  the  given  stim  to  the  time  the  interest 
becomes  due,  and  add  it  to  the  principal.  Then,  find  the  interest  on 
this  amount  as  a  new  principal,  and  add  the  interest  to  it,  as  before. 
Proceed  in  the  same  maimer  for  each  successive  period  when  the  in- 
terest becomes  due  until  the  time  of  settlement. 

Subtract  the  principal  from  the  last  amount,  and  the  remainder  will 
be  the  compound  interest. 

Note.  —  When  partial  payments  have  been  made  on  notes  at  compound  in- 
terest, it  is  customary  to  find  the  amount  of  the  given  principal,  and  from  it  to 
subtract  the  sum  of  the  several  amounts  of  the  indorsements. 

Examples. 

2.  What  is  the  amount  of  $  500  for  3  years  at  compound  in- 
terest ?  Ans.  $  595.508. 

3.  What  is  the  compound  interest  of  $970  for  2  years  9 
months  and  24  days  ?  Ans.  $  173.295. 

4.  What  is  the  compound  interest  of  $  300  for  4  years  6 
months,  at  7  per  cent.  ?  Ans.  $  107.001. 

5.  What  is  the  compound  interest  of  $  316  for  3  years  4 
months  and  18  days?  Ans.  $  69.017. 

377  a  The  computation  of  compound  interest  is  rendered 
more  expeditious  by  means  of  the  following 


COMPOUND   INTEREST. 


281 


TABLE, 


SHOWING   THE  AMOUNT   OF  ONE  DOLLAR  AT   COMPOUND   INTEREST  FOR  ANY 
NUMBER   OF   YEARS  NOT   EXCEEDING  FIFTY. 


No. 
~2 

3  per  cent. 

3£  per 

3ent. 

4  per  cent. 

5  per  cent. 

6  per  cent. 
~i.06(f000 

7  per  cent. 

~1.030~ 

000 

1.035 

000 

1.040 

000 

1.050 

000 

1.070  000 

2 

1.060 

900 

1.071 

225 

1.081 

600 

1.102 

500 

1.123  606 

1.144  900 

3 

1.092 

727 

1.108 

718 

1.124 

864 

1.157 

625 

1.191  016 

1.225  043 

4 

1.125 

509 

1.147 

523 

1.169 

859 

1.215 

506 

1.262  477 

1.310  796  1 

5 

1.159 

274 

1.187 

686 

1.216 

653 

1.276 

282 

1.338  226 

1.402  552 

6 

1.194 

052 

1.229 

255 

1.265 

319 

1.340 

096 

1.418  519 

1.500  730 

7 

1.229 

874 

1.272 

279 

1.315 

932 

1.407 

100 

1.503  630 

1.605  781 

8 

1.266 

770 

1.316 

809 

1.368 

569 

1.477 

455 

1.593  848 

1.718  186 

9 

1.304 

773 

1.362 

897 

1.423 

312 

1.551 

328 

1.689  479 

1.838  459 

10 

1.343 

916 

1.410 

599 

1.480 

244 

1.628 

895 

1.790  848 

1.967  151 

11 

1.384 

234 

1.459 

970 

1.539 

454 

1.710 

339 

1.898  299 

2.104  852 

12 

1.425 

761 

1.511 

069 

1.601 

032 

1.795 

856 

2.012  196 

2.252  192 

13 

1.468 

534 

1.563 

956 

1.665 

074 

1.885 

649 

2.132  928 

2.409  845 

14 

1.512 

590 

1.618 

694 

1.731 

676 

1.979 

932 

2.260  904 

2.578  534 

15 

1.557 

967 

1.675 

349 

1.800 

944 

2.078 

928 

2.396  558 

2.759  032 

16 

1.604 

706 

1.733 

986 

1.872 

981 

2.182 

875 

2.540  352 

2.952  164 

17 

1.652 

848 

1.794 

675 

1.947 

901 

2.292 

018 

2.692  773 

3.158  815 

18 

1.702 

433 

1.857 

489 

2.025 

817 

2.406 

619 

2.854  339 

3.379  932 

19 

1.753 

506 

1.922 

501 

2.106 

849 

2.526 

950 

3.025  600 

3.616  526 

20 

1.806 

111 

1.989 

789 

2.191 

123 

2.653 

298 

3.207  135 

3.869  684 

21 

1.860 

295 

2.059 

431 

2.278 

768 

2.785 

963 

3.399  564 

4.140  562 

22 

1.916 

103 

2.131 

512 

2.369 

919 

2.925 

261 

3.603  537 

4.430  402 

23 

1.973 

587 

2.206 

114 

2.464 

716 

3.071 

524 

3.819  750 

4.740  530 

24 

2.032 

794 

2.283 

328 

2.563 

304 

3.225 

100 

4.048  935 

5.072  367 

25 

2.093 

778 

2.363 

245 

2.665 

836 

3.386 

355 

4.291  871 

5.427  433 

26 

2.156 

591 

2.445 

959 

2.772 

470 

3.555 

673 

4.549  383 

5.807  353 

27 

2.221 

289 

2.531 

567 

2.883 

369 

3.733 

456 

4.822  346 

6.213  868 

28 

2.287 

928 

2.620 

177 

2.998 

703 

3.920 

129 

5.111  687 

6.648  838 

29 

2.356 

566 

2.711 

878 

3.118 

651 

4.116 

136 

5.418  388 

7.114  257 

30 

2.427 

262 

2.806 

794 

3.243 

398 

4.321 

942 

5.743  491 

7.612  255 

31 

2.500 

080 

2.905 

031 

3.373 

133 

4.538 

039 

6.088  101 

8.145  113 

32 

2.575 

083 

3.006 

708 

3.508 

059 

4.764  941 

6.453  387 

8.715  271 

33 

2.652 

335 

3.111 

942 

3.648 

381 

5.003 

189 

6.840  590 

9.325  340 

34 

2.731 

905 

3.220 

860 

3.794 

316 

5.253 

348 

7.251  025 

9.978  114 

35 

2.813 

862 

3.333 

590 

3.946 

089 

5.516 

015 

7.686  087 

10.676  581 

36 

2.890 

278 

3.450 

266 

4.103 

933 

5.791 

816 

8.147  252 

11.423  942 

37 

2.985 

227 

3.571 

025 

4.268 

090 

6.081 

407 

8.636  087 

12.223  618 

38 

3.074 

783 

3.696 

011 

4.438 

813 

6.385 

477 

9.154  252 

13.079  271 

39 

3.167 

027 

3.825 

372 

4.616 

366 

6.704 

751 

9.703  507 

13.994  820 

40 

3.262 

038 

3.959 

260 

4.801 

021 

7.039  989 

10.285  718 

14.974  458 

41 

3.359 

899 

4.097 

834 

4.993 

061 

7.391 

988 

10.902  861 

16.022  670 

42 

3.460 

696 

4.241 

258 

5.192 

784 

7.761 

588 

11.557  033 

17.144  257 

43 

3.564 

517 

4.389 

702 

5.400 

495 

8.149 

667 

12.250  455 

18.344  355 

44 

3.671 

452 

4.543 

342 

5.616 

515 

8.557 

150 

12.985  482 

19.628  460 

45 

3.781 

596 

4.702 

358 

5.841 

176 

8.985 

008 

13.764  611 

21.002  452 

46 

3.895 

044 

4.866 

941 

6.074 

823 

9.434 

258 

14.590  487 

22.472  623 

47 

4.011 

895 

5.037 

284 

6.317 

816 

9.905 

971 

15.465  917 

24.045  707 

48 

4.132 

252 

5.213 

589 

6.570 

528 

10.401 

270 

16.393  872 

25.728  907 

49 

4.256 

219 

5.396 

065 

6.833 

349 

10.921 

333 

17.377  504 

27.529  930 

50 

4.383 

906 

5.584 

927 

7.106 

683 

11.467 

400 

18.420  154 

29.457  025 

Note. —If  each  of  the  numbers  in  the  table  be  diminished  by  1,  the  re- 
mainder will  denote  the  interest  of  $  1,  instead  of  its  amount. 

24* 


282  COMPOUND   INTEREST. 

Ex.  1.     What  is  the  compound  interest  of  $  360  for  5  years 
6  months  and  24  days  ?  Ans.  $  138.14. 

OPERATION. 

Amount  of  $  1  for  5  years,  $1.3  38226 

Principal,  3  6  0 

8  0  2*93560 
40146780 


Amount  of  $  360  for  5  years,  4  8  1.7  6  1  3  6  0 

Amount  of  $  1  for  6mo.  24d.,  1.0  3  4 

1927045440 

1445284080 
481761360 


Amount  of  $  360  for  5y.  6mo.  24d.,  49  8.1  41246240 

Principal,  3  6  0. 

Comp.  int.  of  S  360  for  5y.  6mo.  24d.,      $  1  3  8.1  4  Ans. 

We  find  the  amount  of  S  1  for  5  years  in  the  table,  and,  multiplying 
it  and  the  number  denoting  the  given  prineipal  together,  obtain  the 
amount  of  the  $  360  for  5  years.  On  this  amount  as  a  new  principal 
we  find  the  amount  for  the  remaining  6  months  and  24  days,  tyy  mul- 
tiplying by  the  number  denoting  the  amount  of  $  1  for  the  same  time. 
From  the  last  amount  subtracting  the  original  prineipal,  we  have  left 
the  compound  interest  required.     Hence, 

Multiply  the  am&unt  of  $  1  for  the  given  time  and  rate,  as  found  in 
the  table,  by  the  number  ai  noting  the  given  principal.  The  product  tvill 
be  the  required  amount,  from  which  subtract  the  given  principal,  and  the 
remainder  icill  be  the  compound  interest. 

Note.  —  When  the  given  time  includes  not  only  the  regular  periods  at  which 
interest  becomes  due,  but  also  a  partial  period,  as  a  succession  of  periods  of  a 
year  each,  followed  by  one  containing  months  or  days,  or  both,  after  finding 
the  amount  for  the  regular  periods,  multiply  that  amount  by  the  amount  of  $  1 
for  the  remaining  time  or  partial  period,  and  the  product  will  be  the  required 
amount  for  the  given  time.  In  like  manner,  when  the  number  of  successive  pe- 
riods exceeds  the  limits  of  the  table,  make  the  computations  for  a  convenient 
length  of  time  by  means  of  the  table,  and  on  the  amount  thus  found  make 
another  computation  by  means  of  the  table,  and  so  on. 

In  making  computation  for  a  succession  of  periods  shorter  or  longer  than  one 
year  each,  use  the  numbers  in  the  table  the  same  as  if  the  periods  were  those 
of  one  year  each. 

Examples. 

2.  What  is  the  compound  interest  of  $  1200  for  11  years  at 
7  per  cent.  ?  Ans.  $  1325.822. 


COMPOUND   INTEREST.  283 

3.  What  is  the  compound  interest  of  $  300  for  10  years  7 
months  and  15  days  ?  Ans.  $  257.401. 

4.  What  is  the  compound  interest  of  $  5  for  50  years  at  7 
percent.?  Ans.  $142,285. 

5.  What  is  the  amount  of  $  480  for  40  years,  at  compound 
interest?  Ans.  $4937.144. 

6.  What  is  the  compound  interest  of  $  40  for  4  years,  at  7 
percent.?  Ans.  $12,431. 

7.  What  is  the  compound  interest  of  $  100  for  100  years? 

Ans.  $  33830.20. 

8.  What  is  the  difference  between  the  simple  and  the  com- 
pound interest  of  $  1000  for  33  years  and  4  months  ? 

9.  To  what  sum  will  $  50,  deposited  in  a  savings  bank, 
amount  to  at  compound  interest  for  21  years,  at  3  per  cent., 
payable  semiannually?  Ans.  $173,034. 

(10.)     $  100.  Boston,  September  25,  1853. 

I    For  value  received,  I  promise  to  pay  J.  D.  Forster,  or  order, 
on  demand,  one  hundred  dollars,  with  interest,  after  six  months. 

Allen  T.   Dawes. 

On  this  note  are  the  following  indorsements :  —  June  11, 1854,  received  fifty 
dollars;  September  25,  1854,  received  fifty  dollars. 

What  was  due,  reckoning  at  compound  interest,  August  25, 
1855  ?  Ans.  $  2.247. 

(11.)     $  1000.  St.  Paul,  January  1,  1850. 

For  value  received,  I  promise  to  pay  Stephen  Howe,  or  bearer, 
on  demand,  one  thousand  dollars,  with  interest  at  7  per  cent. 

Wilson  Goodhue. 

Indorsements:  —  June  10,  1850,  seventy  dollars;  September  25, 1851,  eighty 
dollars;  July  4, 1852,  one  hundred  dollars;  November  11,  1853,  thirty  dollars; 
June  5,  1854,  fifty  dollars. 

At  7  per  cent,  compound  interest,  what  remains  due  April  1, 
1855  ?  Ans.  $  1022.34. 

378.  To  find  the  principal,  the  compound  interest,  the 
time,  and  the  rate  being  given. 

Ex.  1.  What  principal  at  6  per  cent,  compound  interest 
will  produce  $  2370  in  10  years  ?  Ans.  $  3000. 


284  COMPOUND   INTEREST. 

operation.  "VYe  find  the  compound  in- 

$  2370  -J-  .790  =  $  3000  Ans.     terest  of  S  1  for  the  given  time, 

and  at  the  given  rate ;  and  pro- 
ceed as  in  like  cases  in  simple  interest  (Art.  359). 

Rule. — Divide  the  given  compound  interest  by  the  compound  in- 
terest of  $  1  for  the  given  time  at  the  given  rate. 

Examples. 

2.  What  principal,  at  7  per  cent,  compound  interest,  will 
produce  $  205.90  in  6  years  and  6  months  ?    Ans.  $  372.16. 

3.  What  sum  of  money,  at  compound  interest,  will  produce 
$  1026.54  in  3  years  2  months  and  12  days  ?      Ans.  $  5000. 

4.  What  sum  of  money  must  be  invested  at  compound  in- 
terest at  a  semiannual  rate  of  3£  per  cent,  to  produce  $  857.25 
in  154-  years  ?  Ans.  $  450. 

379t  To  find  the  rate  per  cent.,  the  principal,  the  inter- 
est, or  the  amount,  and  the  time  being  given. 

Ex.  1.  At  what  rate  per  cent,  must  $  500  be  at  compound 
interest  to  become  $  703.55  in  7  years  ?        Ans.  5  per  cent. 


OPERATION. 


$703.55  -T-  500  a=  $1.4071,  which  for  7  years,  in  the  table, 
denotes  a  rate  of  5  per  cent. 

Since  $  500  becomes  $  703.55  in  7  years  at  the  required  rate,  $  1 
in  the  same  time  at  the  same  rate  will  amount  to  -g-J^  as  much,  or 
$  1.4071.  Corresponding  to  this  amount  of  $  1  for  the  given  time, 
we  find  in  the  table  (Art.  377)  5  per  cent,  the  rate  required. 

Rule. —  Divide  the  amount  by  the  principal,  and  the  quotient  will  be 
the  amount  of  $  1  for  the  given  time  and  the  required  rate  ;  and  in  the 
table,  over  this  amount,  may  be  found  the  rate  per  cent,  required. 

Note.  —  If  the  given  time  contains  a  part  over  an  exact  number  of  periods, 
look  in  the  table,  against  the  number  denoting  the  whole  periods  or  years  in 
the  given  time,  for  that  amount  of  $  1  which  comes  the  nearest  to  the  one 
found  by  dividing.  Then  see  if  the  approximate  amount,  increased  by  its  rate 
of  interest  for  the  fractional  period,  will  equal  the  other  amount;  if  so,  the  rate 
corresponding  to  the  approximate  amount  will  be  the  rate  per  cent,  required ; 
if  not,  the  rate  of  the  approximate  amount  will  be  as  much  greater  or  smaller 
than  the  required  rate,  as  the  interest  added  to  the  approximate  amount  is 
greater  or  smaller  than  that  required  to  produce  the  amount  found  by  dividing. 
The  rule  can  only  be  well  applied  when  the  rate  per  cent,  sought  is  within  the 
limits  of  the  table. 


COMPOUND    INTEREST.  285 

Examples. 

2.  At  what  rate  per  cent,  will  $  400  amount  to  $  640.405, 
at  compound  interest,  in  12  years  ?  Ans.  4  per  cent. 

3.  At  what  rate  per  cent,  must  $  2500  be  loaned,  to  produce 
$  2096.147  compound  interest,  in  9  years  ?    Ans.  7  per  cent. 

4.  At  what  rate  per  cent,  will  any  sum  of  money  double 
itself  at  compound  interest  in  11T8^  years  ?    Ans.  6  per  cent. 

5.  At  what  rate  per  cent,  will  $  10,000  amount  to  $  31479.70 
in  19  years  and  8f  months  ?  Ans.  6  per  cent. 

380.  To  find  the  time,  the  principal,  the  compound  interest, 
and  the  rate  per  cent,  being  given. 

Ex.  1.  In  what  time  will  $500,  at  7  per  cent,  compound 
interest,  amount  to  $  655.398  ?  Ans.  4  years. 

OPERATION. 

$655,398  —  500  ==  $1.310796,  which,  at  the  given  rate  in 
the  table,  denotes  4  years'  time. 

Since  $500  amounts  to  $655,398  at  7  percent,  in  the  required 
time,  $  1  at  the  same  rate,  in  the  same  time,  will  amount  to  -g^  as 
much,  or  $1.310796.  Corresponding  to  this  amount  of  $  1  at  the 
given  rate,  we  find  in  the  table  (Art.  377)  4  years,  the  time  re- 
quired. 

Rule.  —  Divide  the  amount  by  the  principal,  and  the  quotient  will  he 
the  amount  of  $1  at  the  given  rate  for  the  required  time ;  and  in  the 
table,  against  this  amount,  may  be  found  the  time  required. 

Note.  —  If  the  required  time  cannot  be  found  exactly  in  the  table,  the 
number  against  that  amount  of  $  1  which  under  the  given  rate  is  next  less 
than  the  amount  found  by  dividing,  will  denote  the  whole  periods  or  years. 
Then,  find  the  fractional  period  or  part  of  a  year,  by  dividing  1  whole  period 
or  year  by  the  ratio  of  the  difference  between  the  amount  corresponding  to  the 
whole  periods  or  years  and  that  found  by  dividing  to  the  difference  between 
the  former  of  the  amounts  and  that  next  larger  in  the  table ;  and  the  value  of 
the  fraction  obtained  as  the  result  may  be  expressed  in  months  or  days,  or 
both. 

Examples. 

2.  In  what  time  will  $  400  amount  to  $  640.405  at  4  per 
cent,  compound  interest?  Ans.  12  years. 

3.  In  what  time  will  $  6000  amount  to  $  9021.78  at  7  per 
cent,  compound  interest  ? 


286  DISCOUNT   AND   PRESENT   WORTH. 

4.  In  what  time  will  any  sum  double  itself  at  5  per  cent, 
compound  interest?  Ans.  14y.  2mo.  13d. 

5.  In  what  time  will  any  sum  double  itself  at  6  per  cent, 
compound  interest?  Ans.  lly.  lOmo.  20-)-d. 

6.  A  gentleman  has  deposited  $  450,  for  the  benefit  of  his 
son,  in  a  savings  bank,  at  compound  interest  at  a  semiannual 
rate  of  3£  per  cent.  He  is  to  receive  the  amount  as  soon  as 
it  becomes  $  1781.66^-.  Allowing  that  the  deposit  was  made 
when  the  son  was  1  year  old,  what  will  be  his  age  when  he  can 
come  in  possession  of  the  money  ?  Ans.  21  years. 


DISCOUNT  AND  PRESENT  WORTH. 

381.  Discount  is  an  allowance  or  deduction  made  for 
paying  money  before  it  is  due. 

382.  The  present  worth  is  the  amount  of  ready  money 
that  will  satisfy  a  debt  before  it  is  due.  It  is  equivalent  to  a 
principal  which,  being  put  at  interest,  will  amount  to  the  debt 
at  the  time  of  its  becoming  payable.  Thus,  $  100  is  the  present 
worth  of  $  106  due  one  year  hence  at  6  per  cent. ;  for  $  100 
at  6  per  cent,  will  amount  to  $  10G  in  that  time ;  and  $  6  is  the 
discount. 

383.  In  discount,  the  rate  per  cent.,  the  time,  and  the  sum 
on  which  the  discount  is  made,  are  given,  to  find  the  present 
ivorth,  which  corresponds  precisely  to  the  rate  per  cent.,  the 
time  and  the  amount  being  given,  in  either  simple  or  corn- 
pound  interest,  to  find  the  principal. 

384.  The  interest  or  percentage  of  any  sum  cannot  prop- 
erly be  taken  for  the  true  discount ;  for  we  have  seen  (Art. 
382)  that  the  interest  for  one  year  is  the  fractional  part  of  the 
sum  at  interest,  denoted  by  the  rate  for  the  numerator,  and  100 
for  the  denominator ;  and  the  discount  for  one  year  is  the  frac- 
tional part  of  the  sum  on  which  discount  is  to  be  made,  denoted 
by  the  rate  for  the  numerator,  and  100  plus  the  rate  for  the  de- 


DISCOUNT    AND    PRESENT   WORTH.  287 

nominator.  Thus,  if  the  rate  of  interest  is  6  per  cent.,  the 
interest  for  one  year  is  yg^  of  the  sum  at  interest;  but  if  the 
rate  per  cent,  of  discount  is  6,  the  discount  for  one  year  is  T§^ 
of  the  sum  on  which  discount  is  made. 

385.  Business  men,  however,  often  deduct,  or  "take  off," 
from  the  face  of  a  bill  or  note  due  at  some  future  time,  a 
greater  percentage  than  the  interest  would  be  for  the  given 
time  at  the  given  rate.  Therefore,  the  true  present  worth  and 
discount  are  not  obtained  by  that  method,  but  only  a  nominal 
present  worth  and  a  nominal  discount.  The  true  discount 
is  equal  to  the  interest  on  the  true  present  worth  of  the  debt, 
while  the  nominal  discount  is  equal  to  the  interest  on  the  face 
of  the  debt. 

386.  To  find  the  true  present  worth  of  any  sum,  and  the 
discount,  for  any  time,  at  any  rate  per  cent. 

Ex.  1.  What  is  the  present  worth  of  $  12.72,  due  one  year 
hence,  discounting  at  6  per  cent.  ?     What  is  the  discount  ? 

Ans.  $  12  present  worth  ;  $  0.72  discount. 


OPERATION. 


Amount  of  $  1,        1.0  6  )  $  1  2.7  2  (  $  1  2,  Present  worth. 

106 


2  12     $  1  2.7  2,  Given  sum. 
2  12        1  2.0  0,  Present  worth. 

$  0.7  2,  Discount. 

Since  $  1  is  the  present  worth  of  $  1.06  due  one  year  hence,  at  6 
per  cent.,  it  is  evident  that  the  present  worth  of  $12.72  must  be  as 
many  dollars  as  $  1.06  is  contained  times  in  $  12.72.  We  thus  find 
the  present  worth  to  be  $  12,  which,  subtracted  from  the  given  sum, 
gives  $  0.72  as  the  discount. 

Rule.  —  Divide  the  given  sum  by  the  amount  of  $  1  for  the  given 
time  and  rate,  and  the  quotient  will  be  the  present  worth. 

From  the  given  sum  subtract  the  present  worth,  and  the  remainder  will 
be  the  discount. 

Examples. 

2.  What  is  the  discount  on  $  802.50,  at  7  per  cent.,  due  one 
year  hence  ?  Ans.  $  52.50. 

3.  What  is  the  present  worth  of  $117.60,  due  one  year 
hence,  at  12  per  cent.  ? 


288  DISCOUNT   AND   PRESENT   WORTH. 

4.  What  is  the  present  worth  of  $769.60,  due  3  years  and 
5  months  hence  ?  Ans.  $  638.672. 

5.  What  is  the  present  worth  of  $  678.75,  due  3  years  7 
months  hence,  at  7  J  per  cent.  ?  Ans.  $  534.975. 

6.  What  is  the  discount  on  $  600,  due  5  years  hence,  at  5 
per  cent.  ? 

7.  A  merchant  has  given  two  notes  ;  the  first  for  $  79.87,  to 
be  paid  JaAuary  21,  1856;  the  second  for  $87.75,  to  be  paid 
December  17,  1856.  How  much  ready  money  will  discharge 
both  notes  February  10,  1855  ?  Ans.  $  154.545. 

8.  C.  Gardner  owes  Samuel  Hall  as  follows:  $365.87,  to  be 
paid  December  19,  1855;  $161.15,  to  be  paid  July  16,  1856; 
$  112.50,  to  be  paid  June  23,  1854  ;  $  96.81,  to  be  paid  April 
19,  1858.  What  should  Hall  receive  as  an  equivalent,  January 
1,  1854  ?  Ans.  $  653.40. 

9.  What  is  the  present  worth  of  $  67.25  due  3  years 
hence  ? 

10.  What  is  the  present  worth  of  $80,095,  due  3  years 
hence  at  compound  interest  ?  Ans.  $  67.25. 

11.  What  is  the  discount  on  $  110.364  due  5  years  hence,  at 
7  per  cent,  compound  interest  ?  Ans.  S  31.677. 

387.  To  find  a  nominal  present  worth  of  any  sum  due  at 
some  future  time,  and  the  discount  on  the  same  at  a  given 
rate,  reckon  the  interest  on  the  face  of  the  debt  for  the  given 
time  and  rate,  and  the  same  will  be  the  nominal  discount ;  and 
this  discount  subtracted  from  the  face  of  the  note  will  give  the 
nominal  present  worth. 

Examples. 

1.  I  have  bought  of  Paine  and  Woodard  a  bill  of  goods 
amounting  to  $960,  on  six  months,  but  for  ready  money  they 
take  off  from  the  face  of  the  bill,  for  the  time,  5  per  cent. 
What  was  the  amount  paid  ?  '  Ans.  $  912. 

2.  How  much  more  is  the  nominal  than  the  true  discount 
on  $  5000  due  one  year  hence,  at  7  per  cent.  ?  } 

3.  When  money  is  worth  6  per  cent,  a  year,  how  much  may 
be  gained  by  hiring  money  to  pay  $  4440  due  6  months  hence, 
allowing  the  present  worth  of  this  debt  to  be  reckoned  by  de- 
ducting the  nominal  discount?  Ans.  $3,996. 


BANKING.  289 


BANKING. 


388.  Banking  is  the  general  business  transacted  at  banks. 
A  bank  is  a  joint-stock  company,  established  for  the  purpose 

of  receiving  deposits,  loaning  money,  dealing  in  exchange,  or 
issuing  bank-notes  or  bills,  as  a  circulating  medium,  redeemable 
in  specie  at  its  place  of  business.  ., 

The  capital  of  a  bank  is  the  money  paid  in  by  its  stockholders, 
as  the  basis  of  business. 

The  affairs  of  a  bank  are  usually  managed  by  a  board  of  di- 
rectors chosen  by  the  stockholders,  and  the  principal  officers 
are  a  president,  a  cashier,  and  one  or  more  tellers. 

Note.  —  The  president  and  cashier  sign  the  bills  issued ;  the  cashier  super- 
intends the  bank  accounts;  and  the  tellers  receive  and  pay  out  money.  A 
bank  check  is  an  order  drawn  on  the  cashier  for  money. 

389.  Bank  discount  is  the  simple  interest  of  a  note,  draft, 
or  bill  of  exchange,  deducted  from  it  in  advance,  or  before  it 
becomes  due. 

The  interest  is  computed,  not  only  for  the  specified  time,  but 
for  three  days  additional,  called  days  of  grace. 

The  legal  rate  of  discount  is  usually  the  same  as  the  legal 
rate  of  interest ;  and  the  difference  between  bank  discount  and 
true  discount  is  the  same  as  the  difference  between  interest  and 
true  discount  (Art.  384). 

390.  A  note  is  said  to  be  discounted  at  a  bank,  when  it  is 
reeeived  as  security  for  the  money  that  is  paid  for  it,  after  de- 
ducting the  interest  for  the  time  it  was  given. 

The  money  paid  for  a  note  is  called  its  avails,  proceeds,  or 
present  worth. 

391.  A  note,  though  nominally  due  at  the  time  specified  in 
it,  is  not  legally  due  till  the  days  of  grace  have  been  counted. 

The  time  a  note  has  to  run  is  counted  in  days  from  after  the 
day  of  its  being  discounted  to  the  day  of  its  becoming  legally 
due. 

392.  To  find  the  bank  discount  and  the  avails  or  proceeds 
of  a  note  or  bill  for  any  time  or  rate  per  cent. 

N       25 


290  BANKING. 

Ex.  1.  What  is  the  discount  on  a  note  for  $1000  having 
63  days  to  run,  discounted  at  a  bank  at  6  per  cent.  ?  How 
much  are  the  proceeds  ? 

Ans.    Discount,  $  10.50  ;  proceeds,  $  989.50. 

operation.  wre  find  the  interest 

Sum  discounted,  $  1  0  0  0.0  0     on  the  sum  discounted 

,       r  •   .    n      nr\  i  -i  a  a  a     as  in  Art.  354,  and  this 

T^  of  sum  =  int.  for  60d.  1  0.0  0     interest   b  ^e  bank 

2*5  of  int.  for  GOd.  =  int.  for  3d.         .5  0     discount   (Art.    389), 

Bank  discount,  $10.5  0     ^hich  subtracted  from 

_ the     sum    discounted 

Proceeds,  or  present  worth,    $  9  8  9.5  0     gives  the  proceeds  or 

present  worth.  Hence, 

Find  the  interest  on  the  note,  or  sum  discounted,  for  the  given  rate 
and  time,  including  three  days  of  grace,  and  this  interest  will  "be  the 
discount. 

Subtract  the  discount  from  the  face  of  the  note,  or  sum  discounted, 
and  the  remainder  will  be  the  proceeds,  or  present  worth. 

Examples, 

2.  What  is  the  bank  discount  on  a  note  for  $  7800  on  90 
days'  time  ?  Ans.  $  120.90. 

3.  What  is  the  bank  discount  on  a  note  for  $  1200,  payable 
in  GO  days,  at  7  per  cent.  ?  Ans.  $  14.70. 

4.  What  is  the  bank  discount  on  $  8000,  payable  in  60  days  ? 
What  are  the  proceeds  ? 

Ans.  Discount,  $84;  proceeds,  $7916. 

5.  How  much  money  should  be  received  on  a  note  for 
$  760,  payable  in  5  months,  discounted  at  a  bank  ? 

)  &.  A  merchant  sold  a  cargo  of  hemp  for  $7860,  for  which 
he  received  a  note  payable  in  6  months.  How  much  money 
will  he  receive  at  a  bank  for  this  note  ?  Ans.  $  7620.27. 

7.  If  the  following  note  was  discounted  April  3,  1857,  how 
long  had  it  to  run,  and  what  were  the  proceeds  ? 

$  160TV<j.  Boston,  December  3,  1856. 

Six  months  after  date,  for  value  received,  we  promise  to  pay 
Robert  S.  Davis  8?  Co.,  or  order,  one  hundred  and  sixty  T40°0 
dollars,  at  the  Merchants'  Bank, 

Hallett,  Osgood,  &  Co. 
Ans.  63  days;  proceeds,  $158.72. 


BANKING.  291 

8.  Required  the  time  when  the  following  note  will  become 
legally  due,  and  the  bank  discount,  provided  it  was  discounted 
May  16,  1856,  the  rate  being  7  per  cent. 

$89OT50V  Chicago,  April  16, 1856. 

One  hundred  and  twenty  days  after  date,  I  promise  to  pay  to 
the  order  of  Keen  and  Lee  eight  hundred  and  ninety  -f^  dollars, 
at  the  Marine  Bank  Thomas  L.  Cook. 

Ans.  August  17,  1856;  discount,  $16,103. 

9.  Required  the  legal  time  of  maturity  of  the  following  note, 
and  the  proceeds,  it  having  been  'discounted  June  11,  1856. 

$  1340.  Philadelphia,  May  1,  1856. 

For  value  received,  ninety  days  after  date,  we  promise  to  pay 
J.  B.  Lippi?icott  §  Co.,  or  order,  one  thousand  three  hundred 
and  forty  dollars,  at  the  Girard  Bank,  without  defalcation, 

Johnson  &  Pollock. 

10.  The  following  note  was  discounted,  at  2  per  cent,  a 
month,  July  5,  1857  ;  how  long  had  it  to  run,  and  what  were 
the  proceeds  ? 

$  9000.  New  York,  June  19,  1857. 

Two  months  after  date,  for  value  received,  I  promise  to  pay 
to  the  order  of  Joseph  Appleton,  nine  thousand  dollars,  at  the 
Manhattan  Bank.  L.  T.  Roberts. 

Ans.  48  days  ;  proceeds,  $  8712.00. 

393  •  To  find  the  amount  for  which  a  note  must  be  given 
that  the  proceeds  shall  be  a  required  sum. 

Ex.  1.  For  what  amount,  payable  in  60  days,  must  a  note  be 
given  to  a  bank  discounting  at  6  per  cent.,  to  obtain  $  989.50  ? 

Ans.  $  1000. 

operation.  Since  $  0.9895  of  pro- 

$  1.0  0  0  0  ceeds  requires  $  1  to  be 

Interest  of  $  1  for  63  days,      .0105  discounted,  $  989.50  will 

require  as  many  dollars 

Proceeds  of  $  1,                     $  0.9  8  9  5  as  $  0.9895  is  contained 

<£Q89f)0-^   9895—^1000  times     m     $  989.50,     or 

Divide  tlie  given  sum  ~by  the  proceeds  of  $  1  for  tlie  given  time  and 
rate  of  lank  discount,  including  three  days  of  grace,  and  the  quotient 
will  be  the  required  amount. 


292  BANKING. 


Examples. 


2.  What  sum,  payable  in  90  days,  if  discounted  at  7  per 
cent,  bank  discount,  will  produce  $  680  ?  Ans.  $  G92.52. 

3.  What  must  be  the  face  of  a  note,  which,  when  discounted 
at  a  bank  for  120  days,  shall  give  as  its  proceeds  $  540.50  ? 

Ans.  $551.81. 

4.  For  what  amount  must  a  note  be  given,  in  order  that  the 
proceeds  of  the  note  discounted,  when  having  6  months  to  run, 
shall  be  $  1938  ? 

5.  The  avails  of  a  note,  having  4  months  to  run,  discounted 
at  a  bank,  were  $1631.60 ;  what  was  the  face  of  the  note  ? 

Ans.  $1665.74. 

6.  If  a  gentleman  wishes  to  obtain  $  1500,  for  what  sum 
must  he  give  his  note  payable  in  30  days,  allowing  it  is  to  be 
discounted  at  1  per  cent,  a  month  ?  Ans.  $  1516.68. 

394.  To  find  the  rate  of  interest  corresponding  to  a  given 
rate  of  bank  discount. 

Ex.  1.  What  rate  of  interest  is  paid  when  a  note  payable 
in  60  days  is  discounted  at  6  per  cent.  ? 

Ans.  6T^6g-  per  cent. 

operation.  Every  $  1  discounted  for  the  given 

.06  ~  .9895  =  .06TV/y6g-       time  and  rate  yields  as  its  proceeds 

S  0.9895.  Then,  if  $  1  in  the  given 
time  yield  a  certain  interest  at  6  per  cent,  $  0.9895  in  the  same 
time  will  yield  the  same  interest,  at  as  many  per  cent,  as  the  given 
rate,  .06,  contains  times  .9895. 

Rule.  —  Divide  the  given  rate  per  cent,  expressed  decimally,  by  the 
number  denoting  the  proceeds  of  $  1  for  the  given  time  and  rate.  The 
quotient  will  be  the  rate  of  interest  required. 

Examples. 

2.  What  rate  of  interest  is  paid  when  a  note  payable  in  30 
days  is  discounted  at  6  per  cent.  ?  Ans  6Tf  f  F  per  cent. 

3.  What  rate  of  interest  is  paid  when  a  note  payable  in  90 
days  is  discounted  at  6  per  cent.  ?  Ans.  6Ty6%  per  cent. 

4.  A  note  payable  in  4  months  is  discounted  at  2  per  cent,  a 
month ;  what  rate  of  interest  is  paid  ? 

5.  When  a  note,  payable  in  6  months,  is  discounted  at  7  per 


BANKING.  293 

cent.,  to  what  rate  of  interest  does  the  bank  discount  corre- 
spond ? 

6.  When  a  note  payable  in  one  year,  without  grace,  is  dis- 
counted, to  what  rate  per  cent,  of  interest  does  the  bank  discount 
correspond  ?  Ans.  6£f  per  cent. 

395.  To  find  the  rate  of  bank  discount  corresponding  to  a 
given  rate  of  interest. 

Ex.  1.  At  what  rate  of  bank  discount  must  a  note,  payable 
in  60  days,  be  discounted,  to  produce  6  per  cent,  interest  ? 

Ans.   5^-f  |^  per  cent. 
operation.  Since  bank  discount  is  reckoned 

.06  -7-  1.0105  =  .05££££.  on  an  amount  which  equals  the  pro- 
ceeds plus  the  interest  on  the  pro- 
ceeds for  the  given  time  and  rate  (Art.  389),  the  bank  discount  for 
every  $  1  of  proceeds  is  equivalent  to  the  interest  on  $  1  -)-  .0105 
=  $  1.0105.  Then,  if  $  1  yield  a  certain  interest  in  a  given  time  at 
6  per  cent,  per  annum,  $  1.0105  must  yield  the  same  interest  in  the 
same  time,  at  as  many  per  cent,  per  annum  as  the  given  rate,  .06,  con- 
tains times  1.0105. 

Rule.  — Divide  the  given  rate  per  cent,  expressed  decimally,  by  the 
number  denoting  the  amount  of  $  1  for  the  given  time  and  rate.  The 
quotient  will  be  the  rate  of  bank  discount  required. 

Examples. 

2.  At  what  rate  must  a  note,  payable  in  30  days,  be  dis- 
counted, to  produce  6  per  cent,  interest? 

Ans.  5jf^  per  cent. 

3.  At  what  rate  must  a  note,  running  60  days,  be  discounted, 
to  yield  2  per  cent,  a  month  interest?    Ans.  23^^  per  cent. 

4.  At  what  rate  must  a  note,  running  90  days,  be  discounted, 
to  produce  6  per  cent,  interest  ? 

5.  At  what  rate  must  a  note,  running  120  days,  be  dis- 
counted, to  produce  8  per  cent,  interest  ? 

6.  At  what  rate  must  a  note,  running  6  months,  be  dis- 
counted, to  produce  7  per  cent,  interest  ? 

7.  At  what  rate  must  a  note,  payable  1  year  hence,  without 
grace,  be  discounted,  to  produce  6  per  cent,  interest  ? 

8.  What  rate  of  bank  discount  on  a  note,  payable  8  years 
and  4  months  hence,  without  gracefcorresponds  to  5  per  cent. 
interest  ?  Ans.  3/1-  per  cent. 

25* 


294  MISCELLANEOUS   EXAMPLES. 


MISCELLANEOUS  EXAMPLES. 

1.  The  gold  coinage  of  the  United  States  contains  9  parts  of 
pure  gold  to  1  part  of  alloy,  and  the  alloy  is  1  part  copper  to  1 
part  silver.     What  per  cent,  of  the  whole  is  each  metal  ? 

Ans.  Gold  90  per  cent. ;  silver  5  per  cent. ;  copper  5  per  cent. 

2.  How  much  grain  must  be  sent  to  the  miller  that  a 
bushel  of  meal  may  be  returned,  the  miller  taking  y1^  part 
for  toll  ?  Ans.  34T25qts. 

3.  If  a  gentleman,  possessing  $  25000,  has  a  net  yearly  in- 
come of  4  per  cent,  of  that  sum,  how  much  may  he  spend  each 
year  in  order  that  his  expenses  shall  just  equal  his  income  ? 

4.  The  half-dollar  of  coinage  previous  to  1853  contains  20 Gj- 
grains  of  standard  silver,  and  that  of  present  coinage  contains 
192  grains ;  what  per  cent,  more  of  standard  silver  does  the 
one  contain  than  the  other  ?  Ans.  7-gJ. 

5.  My  horse  is  worth  50  per  cent,  more  than  my  buggy ; 
how  many  per  cent,  is  the  buggy  worth  less  than  the  horse  ? 

Ans.  33£. 
G.  How  many  years  longer  will  it  take  $  10  to  become  $  20 
at  5  per  cent,  than  at  G  per  cent,  simple  interest  ? 

Ans.  3 -J  years. 

7.  "What  is  the  present  worth  of  $  500  due  4  years  hence,  at 
5  per  cent,  compound  interest  ?  Ans.  $  411.351. 

8.  Bought  cloth  at  $5.00  per  yard.  What  must  be  "the 
asking  price,"  in  order  to  fall  on  it  10  per  cent.,  and  still  make 
10  per  cent,  on  the  purchase  ?  Ans.  $  G.ll^. 

9.  A  merchant  sold  a  cargo  of  hemp  for  $  7860,  for  which 
he  received  a  note  payable  in  6  months.  How  much  money 
will  he  receive  for  the  note  at  a  bank  ?  Ans.  $  7G20.27. 

10.  What  is  the  difference  between  the  true  discount  and 
that  taken  by  banks  on  $  1500  due  one  year  hence  without 
grace?  Ans.  $5.09§£. 

11.  When  a  note  payable  in  GO  days  is  discounted  at  the 
rate  of  2  per  cent,  a  month,  what  rate  of  interest  is  bank  dis- 
count equal  to?  .   JL  Ans.  25 ^fy  per  cent. 

12.  The  proceeds  of  a  45  day  note  discounted  at  1±  per 
cent,  a  month  yielded  $  3G.40.  What  was  the  face  of  the 
note?  Ans.  $1G17.77£. 


stocks.  295 

13.  At  what  rate  per  cent,  a  month  of  bank  discount  should 
a  30  day  note  be  discounted,  that  interest  may  be  received  at 
the  rate  of  12  per  cent.  ?  Ans.  llfff. 

14.  How  much  more  can  a  bank  make  in  693  days  with 
$50000  by  discounting  notes  on  30  days'  time,  than  by  dis- 
counting those  on  60  days,  the  rate  of  discount  being  6  per 
cent.,  and  the  profits  in  both  cases  to  be  retained  in  the  bank 
till  the  end  of  the  time  ? 

15.  A  merchant  bought  450  quintals  of  fish  at  $  3.50  cash, 
and  sold  them  immediately  for  $  4.00  on  6  months'  credit,  for 
which  he  received  a  note.  If  he  should  get  this  note  discounted 
at  a  bank,  what  will  he  gain  on  the  fish  ?  Ans.  $  170.10. 

1 6.  A  bank  by  discounting  a  note  at  6  per  cent,  receives  for 
its  money  a  discount  equivalent  to  6J  per  cent,  interest.  How 
long  must  the  note  have  been  discounted  before  it  was  due  ? 

Ans.  lyr.  3mo.  12d. 


STOCKS. 

396.  Stocks  is  a  general  name  given  to  government  bonds, 
and  to  money  capital  invested  in  corporations. 

The  capital  of  banks,  and  of  insurance,  railroad,  manufac- 
turing, mining,  andlike  companies,  is  usually  divided  into  equal 
shares,  the  market  Value  of  which  is  often  variable. 

397  •  Stocks  are  said  to  be  at  par  when  they  sell  for  their 
original  value  ;  above  par,  or  at  a  premium,  when  for  more 
than  their  original  value  ;  below  par,  or  at  a  discount,  when  for 
less  than  their  original  value. 

898.  The  premium  and  discount  on  stocks  are  generally 
computed  at  a  certain  per  cent,  on  the  original  or  nominal 
value  of  the  shares. 

Note.  —  The  original  value  of  a  share  of  bank,  insurance,  railroad,  or  like 
stock,  is  usually  $  i00,  but  sometimes  $  50,  and  rarely  any  other  sum  than  one 
of  these. 

$89.     A  dividend  is   the  interest  or  profit  on  stocks,  dis- 


296  stocks. 

tributed  to  the  shareholders,  and  is  reckoned  on  the  par  or 
nominal  value  of  the  shares. 

400.  To  find  the  market  value  of  stocks  when  they  are  at  a 
premium,  or  at  a  discount. 

Ex.  1.  What  is  the  value  of  20  shares  of  bank  stock,  at  9 
per  cent,  premium,  their  nominal  or  par  value  being  $  100 
each  ? 


OPERATION. 


$100  X  20  =  $2000;  $2000  X  1.09  =  $2180. 

Since  the  par  value  of  1  share  is  $  100,  that  of  20  shares  is  $  2000 ; 
then,  as  $  1  at  9  per  cent,  premium  equals  $  1.09,  $  2000  will  equal 
2000  times  as  much,  or  $  2180. 

Rule.  —  Multiply  the  par  value  of  the  given  stock  by  1  increased 
by  the  rate  per  cent,  premium,  or  by  1  decreased  by  the  rate  per  cent. 
discount,  expressed  decimally,  and  the  product  icill  be  the  value  re- 
quired. 

Note.  —  The  difference  between  the  par  and  market  value  gives  the  per 
cent,  of  premium  or  discount. 

Examples. 

2.  What  is  the  value  of  $  24360  of  stock,  at  35  per  cent, 
premium.  Ans.  $  32886. 

3.  Sold  15  shares  of  the  Camden  and  Amboy  Railroad,  the 
par  value  being  $  100  per  share,  at  13  per  cent,  advance.  To 
what  did  they  amount  ?  Ans.  $  1695. 

4.  What  must  be  paid  for  10  shares  of  the  Old  Colony  and 
Fall  River  Railroad,  at  85  per  cent.,  the  original  value  being 
$100  each?    f 

5.  Sold  30  shares,  $  100  each,  in  the  Boston  Bank,  at  8| 
per  cent,  advance.  To  what  did  they  amount,  and  how  much 
was  the  premium  ? 

Ans.  Amount,  $3262.50;  premium,  $262.50. 

6.  What  must  be  given  for  25  shares  of  insurance  stock,  par 
value  being  $50,  at  3  per  cent,  discount?      Ans.  $  1212.50. 

7.  What  must  be  paid  for  22  shares  of  the  Iron  City  Manu- 
facturing Stock,  par  value  being  $  250,  at  95  per  cent.,  and 
how  much  is  the  discount  ?       Ans.  $  5225  ;  discount,  $  275. 

8.  What  will  be  the  cost  of  $  50000  of  United  States  gov- 
ernment stock,  at  17  per  cent,  advance  ?  Ans.  $  58500. 


stocks.  297 

9.  Bought  $  19500  of  State  stocks  at  93  per  cent.,  and  sold 
the  same  at  103  per  cent. ;  how  much  was  gained  by  the  op- 
eration ?  Ans.  $  1950. 

401.  To  find  the  par  value  of  stocks,  when  they  are  at  a 
premium,  or  at  a  discount. 

Ex.  1.  Bought  Ocean  Insurance  Company  stock,  at  7  per 
cent,  premium,  for  $  535  ;  what  is  its  par  value  ?  Ans.  $  500. 

operation.  Since  $  1  at   7  per  cent,  pre- 

$  535  ~  1.07  —  $  500,  Ans.     mium  equals  $  1.07,  the  par  value 

of  the  stock  must  be  as  many  dol- 
lars as  535  contains  times  1.07,  or  $  500. 

Rule.  —  Divide  the  market  value  of  the  given  stock  by  1  increased 
by  the  rate  per  cent,  premium,  or  by  1  decreased  by  ilieWate  per  cent, 
discount,  expressed  decimally,  and  the  quotient  will  be  the  value  re- 
quired. 

Examples. 

2.  Bought  Massachusetts  State  stock,  at  3J  per  cent,  pre- 
mium, for  $  6210 ;  what  is  its  par  value  ?  Ans.  $  6000. 

3.  Sold  11  shares  of  Reading  Bank,  at  5  per  cent,  premium, 
for  $  1155  ;  what  is  the  par  value  of  its  shares  ?   Ans.  $  100. 

4.  Bought  41  shares  of  canal  stock  at  40  per  cent,  below  par 
for  $  1230 ;  what  is  the  par  value  of  its  shares  ?j 

5.  Bought  19  shares  of  bank  stock,  at  a  premium  of  8  per 
cent.,  for  $  2052 ;  what  was  the  amount  of  premium  paid  ? 

6.  When  government  stocks  are  at  5  per  cent,  discount,  how 
much  par  value  will  $  16245  purchase,  and  what  is  amount  of 
discount  ?  Ans.  $  17100  ;  discount,  $  855. 

7.  When  railroad  stock  at  15  per  cent,  advance  is  selling  at 
$57.50  per  share,  how  many  shares  may  be  bought  for  $  862.50, 
and  what  will  be  the  amount  of  premium  ? 

Ans.  15  shares;  $112.50  premium. 

8.  How  many  State  bonds  of  $  1000  each,  at  12  per  cent, 
discount,  can  be  purchased  for  $7920,  and  how  much  may  be 
gained  by  the  operation,  if  the  selling  price  should  afterwards 
advance  to  par?  Ans.  9  bonds;  $  1080  may  be  gained. 

402.  To  find  the  rate  of  interest  to  which  a  dividend  on 
any  stock  bought  at  a  premium  or  discount  corresponds. 


298  stocks.     - 

Ex.  1.  Received  12£  per  cent,  dividend  on  an  investment 
in  stocks  at  25  per  cent,  above  par ;  to  what  rate  per  cent,  in- 
terest did  it  correspond  ?  Ans.  10  per  cent. 

operation.  Since  the  stock  was  bought  at  25 

.125  —  1.25  =  10,  Ans.      per  cent,  above  par,  every  $1.25  of 

investment  must  represent  only  $ 1 
of  par  value.  Then,  since  every  S  1  of  par  value  pays  a  dividend 
corresponding  to  12^  per  cent,  interest,  every  $  1.25  of  investment 
pays  as  many  per  cent,  interest  as  12^  per  cent.  =  .125  is  con- 
tained times  in  1.25,  or  10  per  cent. 

Rule.  —  Divide  the  rate  per  cent,  of  dividend,  expressed  decimal///, 
by  1  increased  by  the  rate  per  cent,  premium,  or  by  1  decreased  by  the 
rate  per  cent,  discount,  expressed  decimally,  and  the  quotient  will  denote 
the  rate  of  interest  required. 

Note.  —  If  it  be  required  to  find  at  what  price  a  stock,  paying  a  certain  rate 
per  cent,  dividend,  should  be  bought  in  order  that  the  investment  shall  pay  a 
given  rate  of  interest,  divide  the  rate  per  cent,  of  dividend,  expressed  decimally, 
by  the  yiven  rate  of  interest,  expressed  decimally,  and  the  quotient  will  be  the  price 
required  of  each  $  1  of  the  yiven  stock. 

Examples. 

2.  Received  G  per  cent,  dividend  on  factory  stock,  purchased 
at  25  per  cent,  below  par.  What  rate  per  cent,  interest  did  the 
investment  pay  ?  Ans.  8  per  cent. 

3.  When  railroad  stock  paying  11  per  cent,  dividend  is 
worth  $110  per  share,  or  $  10  per  share  above  par,  to  what 
rate  of  interest  would  the  income  from  an  investment  in  its 
shares  correspond?  Ans.  10  per  cent. 

4.  How  much  advance  must  be  paid  for  stocks  paying  12  per 
cent,  dividends,  in  order  that  the  investment  shall  pay  exactly 
8  per  cent,  interest  ?  Ans.  50  per  cent. 

5.  Which  is  the  better  investment,  the  buying  of  9  per  cent, 
stocks  at  25  per  cent,  advance,  or  6  per  cent,  stocks  at  25  per 
cent,  discount  ? 

6.  At  what  per  cent,  discount  must  government  5  per  cent, 
stock  be  bought  that  the  investment  may  yield  7  per  cent.  ? 

Ans.  28f-  per  cent. 

7.  How  much  more  income  yearly  may  be  derived  from 
$  20000  invested  in  5  per  cent,  stock  bought  at  20  per  cent, 
discount,  than  by  letting  the  same  sum  at  6  per  cent,  in- 
terest? Ans.  $50. 


BROKERAGE   AND    COMMISSION.  299 


BROKERAGE  AND  COMMISSION. 

403*  Brokerage  is  the  percentage  paid  to  a  dealer  in 
money  and  stocks,  called  a  broker,  for  making  exchanges  of 
money,  negotiating  different  kinds  of  bills  of  credit,  or  transact- 
ing other  like  business. 

404.  Commission  is  the  percentage  paid  an  agent,  factor, 
or  commission  merchant  for  buying  or  selling  goods,  making  col- 
lections, or  transacting  other  business. 

405 1  When  the  person  transacting  the  commission  business 
lives  in  a  foreign  country,  he  is  frequently  called  a  correspond- 
ent or  consignee. 

The  goods  shipped  or  forwarded  to  a  consignee  to  be  sold  on 
commission  are  termed  a  consignment,  and  the  person  sending 
or  consigning  the  same  is  called  the  consignor, 

406.  The  rate  per  cent,  of  brokerage  or  commission  is  not 
regulated  by  law,  but  varies  in  different  places,  and  with  the 
nature  of  the  business  transacted. 

Brokerage  and  commission  are  computed  in  the  same  man- 
ner. 

407.  To  find  the  brokerage  or  commission  on  any  given 
sum. 

Ex.  1.  Paid  a  broker,  for  exchanging  $  896  uncurrent  bills 
for  par  funds,  2  per  cent,  brokerage.  How  much  was  the  bro- 
kerage ?  Ans.  $  17.92. 

Since  brokerage  is  a  percentage  on  the  given  sum,  the  brokerage 
on  %  896  at  2  per  cent,  will  be  $  896  X  .02  ==  $  17.92. 

Rule.  —  Find  the  percentage  on  the  given  sum  at  the  given  rate  per 
cent.,  and  the  result  will  be  the  brokerage  or  commission. 

Note.  —  When  the  brokerage  or  commission,  and  the  sum  on  which  it  is 
reckoned,  are  given,  the  rate  per  cent,  may  be  found  as  in  Art.  348. 

Examples. 

2.  My  agent  in  New  Orleans  has  purchased  cotton,  on  my 
account,  to  the  amount  of  $  18768.  What  is  his  commission 
at  If  per  cent.  ?  Ans.  $  328.44. 


300  BROKERAGE  AND  COMMISSION. 

3.  I  hare  engaged  a  broker  to  purchase  for  me  12  shares  in 
the  New  York  Central  Railroad,  at  $  112.25  per  share;'  what 
is  his  commission  at  ^  per  cent.  ?  Ans.  $  3.3  6|-. 

4.  My  agents,  Hilton  and  Marcy  of  Cincinnati,  advise  me 
that  they  have  purchased  on  my  account  a  cargo  of  pork,  con- 
sisting of  700  barrels,  at  $  12.25  per  barrel ;  what  is  their  com- 
mission at  1-J  per  cent.  ?  Ans.  $  15  0.0  6  J. 

5.  What  rate  per  cent,  of  brokerage  does  a  broker  charge 
who  takes  $  50  for  investing  $  10000  ? 

6.  What  is  the  commission  on  the  sale  of  173cwt.  of  sugar, 
at  $  8.95  per  cwt,  at  1|  per  cent.  ?  Ans.  $  29.03^. 

7.  My  factor  at  Mobile  advises  me  that  he  has  purchased  on 
my  account  37  bales  of  cotton,  at  $  107.75  per  bale ;  what  is 
his  commission  at  §  per  cent.  ?  Ans.  $  14.95^. 

8.  A  consignee  in  London  writes  that  he  has  purchased  for 
his  employer  goods  to  the  amount  of  395£.  15s.  5d. ;  wrhat  is 
his  commission  at  2£  per  cent.?  Ans.  8£.  18s.  lTVu^- 

9.  Paid  G.  Willis  $5.46  for  exchanging  $364  of  depre- 
ciated currency  ;  what  was  the  rate  of  brokerage  ? 

Ans.  li  per  cent. 

408.  When  the  given  amount  includes  both  the  brokerage 
or  commission,  and  the  sum  to  be  invested. 

Ex.  1.  A  gentleman  intrusts  $  20050  to  a  broker  in  New 
York  City,  with  instruction,  after  deducting  his  brokerage  of  £ 
per  cent.,  to  invest  the  balance  in  government  bonds.  What 
will  be  the  sum  invested,  and  how  much  will  be  the  broker- 
age? 

OPERATION. 

$20050-*-$  1.0 025  =  $2000  0,  Investment. 
$20050  —  $20000=  $5  0,  .Brokerage. 

Since  the  broker  is  entitled  to  ^  per  cent,  of  the  sum  he  invests,  it 
is  evident  he  requires  $  1.0025  in  order  to  invest  $  1.  Hence,  the 
investment  he  can  make  will  be  as  many  dollars  as  $  20050  con- 
tains times  1.0025,  or  $  20000 ;  which,  being  subtracted  from  the 
amount  forwarded  him,  leaves  as  his  brokerage  $  50. 

Rule.  —  Divide  the  given  amount  by  1  increased  by  the  rate  per 
cent,  of  brokerage  or  commission,  expressed  decimally,  and  the  quotient 
ivill  be  the  investment. 

Subtract  the  investment  from  the  given  amount,  and  the  remainder  will 
be  the  brokerage  or  commission. 


BROKERAGE   AND    COMMISSION.  301 

NOTE.  —  When  the  brokerage  or  commission  and  the  rate  per  cent,  of  the 
same  are  given,  the  sum  on  which  it  is  reckoned  may  be  found  as  in 
Art.  349. 

Examples. 

2.  An  agent  receives  $  1976,  which  includes  the  sum  he  is 
to  lay  out  in  goods  and  also  his  commission  at  4  per  cent.  How- 
much  of  the  amount  is  to  be  expended  for  the  goods,  and  how 
much  is  his  commission  ? 

Ans.  $  1900  for  the  goods  ;  $  76  commission. 

3.  A  broker  receives  $  8341.50,  which  includes  a  sum  to  be 
invested  in  railroad  shares  at  $  83  each,  and  his  brokerage  at 
£  per  cent.  How  many  shares  can  he  purchase,  and  how  much 
is  his  brokerage  ?         Ans.  100  shares;  $41.50  brokerage. 

4.  Sent  to  my  agent,  John  Crowell,  Eochester,  $8960,  to 
purchase  a  quantity  of  flour ;  his  commission  is  2  per  cent,  on 
the  purchase,  which  he  is  to  deduct  from  the  money ;  what  is 
his  commission  ?  Ans.  $  175.68f  \ . 

5.  A  town  has  levied  a  tax  of  $5150,  which  sum  includes 
the  amount  voted  for  the  repairs  of  a  bridge  and  the  collec- 
tor's commission  of  3  per  cent.  How  much  was  voted  for  the 
bridge,  and  how  much  does  the  collector  receive  for  his  com- 
mission ?         Ans.  $  5000  for  the  bridge ;  $  150  commission. 

6.  What  amount  of  money  has  been  invested  when  the  bro- 
ker's charges,  at  1  J-  per  cent,  for  making  the  investment,  amount 
to  $  285  ? 

7.  A  commission  merchant  purchases  for  me  in  New  Orleans 
34  boxes  of  sugar,  pays  for  cartage  and  freight  $  7.50,  and  his 
commission  is  1 J  per  cent,  on  the  amount  of  purchase,  making 
the  whole  bill  $  740.83  £.  How  much  was  his  commission,  and, 
allowing  250  pounds  to  a  box,  how  much  a  pound  did  he  pay 
for  the  sugar  ?       Ans.  $  10.83f  jcommission ;  $  0.08£  per  lb. 

8.  Sent  a  cargo  of  flour  to  Liverpool,  which  my  factor  sold 
for  987£.  18s.  6d.  He  invested  this  sum  in  broadcloths,  at  l£. 
3s.  8d.  per  yard.  His  commission  for  selling  the  flour  is  2£ 
per  cent.,  and  for  purchasing  the  broadcloth  If  per  cent;,  and 
he  is  to  receive  his  commissions,  for  selling  and  buying,  out  of 
the  proceeds  of  the  flour.  Required  the  number  of  yards  of 
broadcloth  that  I  should  receive.  Ans.  SOl-^fy^yd. 

N      26 


302 


ACCOUNT  OP  SALES. 


ACCOUNT   OF   SALES. 

409t  An  Account  of  Sales  is  an  account  of  goods  sold,  which 
a  commission  merchant  or  consignee,  &c.  furnishes  to  his  em- 
ployer. It  contains  the  quantity  and  price  of  the  goods  dis- 
posed of,  charges  attending  the  sales,  and  the  net  proceeds,  or 
the  sum  to  which  the  owner,  or  consignee,  is  entitled  after  all 
charges  are  deducted. 

The  names  of  purchasers  are  often  included  in  the  account 
of  sales,  as  in  Example  1,  and  sometimes  the  mark  of  the  box, 
bale,  &c,  as  in  Example  2. 

410.  The  gross  amount  of  sales  is  the  sum  of  all  the  quan- 
tities at  the  prices  given ;  and  the  net  proceeds  are  found  by  de- 
ducting the  commission  at  the  given  rate,  and  the  other  charges, 
from  the  gross  amount 

Find  the  commission  and  net  proceeds  required  in  the  fol- 
lowing 

Examples. 

1.  Account  of  sales  of  flour  received  by  the  steamer  Calvert, 
from  Baltimore,  sold  on  account  of  James  Taylor,  Ellicott's  Mills. 


Date. 


1857. 
Jan.  1. 

u 
u 

Jan.  3. 

Feb.  1. 

"     5. 


To  whom  sold. 


William  Hooper, 
Sidney  Pope, 
Albert  Rollo, 
S.  S.  Coe, 
J.  C.  Hill  &  Co., 
G.  T.  Sampson, 


Description. 


Superfine, 
Extra  Eagle, 

U  (« 

Baltimore, 
Extra  Eagle, 

Charges. 


Barrels. 

Price. 

100 

8.00 

50 

7.50 

25 

7.50 

30 

7.00 

CO 

7.00 

25 

7.50 

Commission  on  $  2144.50,  at  1\  per  cent,  $  48.25 
Freight  and  drayage,  41.63 

4.50 


Advertising, 


$  2144.50 


$94.38 
$2050.12 


Net  proceeds  to  credit  of  James  Taylor, 
Errors  excepted. 
Boston,  May  15, 1857.  Niles,  Marvin,  &  Co. 


PROFIT   AND   LOSS. 


303 


2.  Sales  of  goods,  made  by  Haskell,  Fargo,  &  Co.,  on  account 
of  Jones,  Boker,  &  Co.,  New  York. 

Description. 


Date. 

Marks. 

1857. 

Feb.  7. 

A. 

a 

B.  3. 

a 

Z. 

Mar.  3. 

X. 

u 

19. 

Apr.  11. 

Y.  Z. 

"     12. 

L.  L. 

u.     12. 

4.  10. 

4  casks  Sperm  Oil,  Fall, 
13     M  "        u    ^Vinter 
23     "    Whale  Oil,  Crude,  ' 
20  bbls.  Olive  Oil, 

10    "     Turpentine,  Spirits, 
10    "     Varnish, 

5  "     Olive  Oil, 
2    "     Varnish, 


Charges. 

Feb.  19.     Freight,  40  casks  (a)  %  2.50, 
Mar.  20.  "        30  bbls.   "      2.00, 

Apr.  29.  "        17    "       "      1.50, 


Commission,  3  per  cent.,  on  $ 
Cartage,  cooperage,  &c, 
Storage  and  insurance, 


Gals. 

Price. 

208 

1.50 

650 

1.40 

874 

.70 

320 

1.20 

315 

.60 

315 

.20 

157-1 

1.20 

63 

.20 

26.00 
63.24 


Net  proceeds  due  J.,  B.,  &  Co.,  $ 

Errors  and  omissions  excepted. 
Chicago,  August  7,  1857.  Haskell,  Fargo,  &  Co. 


PROFIT  AND   LOSS. 


41 1#  Profit  and  Loss  is  the  process  by  which  merchants 
and  others  estimate  their  gains  or  losses  in  business  transac- 
tions. 

Gains  and  losses  are  usually  reckoned  on  the  prime  or  first 
cost  of  articles. 

412#  To  find  the  selling  price  when  the  cost  and  the  gain 
or  loss  per  cent,  are  given. 

Ex.  1.  If  I  buy  cloth  at  $  4  per  yard,  for  how  much  per 
yard  must  I  sell  it  to  gain  25  per  cent.  ?  Ans.  S  5. 


304  PROFIT   AND   LOSS. 

Since  the  selling  price  is  to  be  a  gain 

operation.  on  the  cost  of  25  per  cent,  evidently  it 

$4  X  1.25  =  $5,  Ans.     must  be  125  per  cent,  of  the  cost,  or 

$4  X  1.25. 

Rule.  — Multiply  the  cost  price  by  1  increased  by  the  gain  per  cent., 
or  by  1  decreased  by  the  loss  per  cent.,  expressed  decimally,  and  the 
product  will  be  the  selling  price  required. 

Examples. 

2.  If  I  buy  cloth  at  $  5  per  yard,  for  what  must  I  dispose  of 
it  per  yard  to  lose  20  per  cent.  ?  Ans.  $  4. 

3.  Bargained  for  cheese  at  $  8.50  per  cwt.  How  must  it  be 
sold  to  gain  10  per  cent.  ?  Ans.  $  9.35  per  cwt. 

4.  Molasses  having  been  bought  at  42  cents  a  gallon,  and  not 
proving  so  good  as  expected,  was  sold  at  a  loss  of  5  per  cent. 
For  what  was  it  sold  a  gallon  ? 

5.  Bought  a  house  and  lot  for  $  2500  ;  at  what  price  must  it 
be  sold  to  gain  20  per  cent.  ?  Ans.  $  3000. 

413*  To  find  the  cost  when  the  selling  price  and  the  gain 
or  loss  per  cent,  are  given. 

Ex.  1.  If  I  sell  cloth  at  $  5  per  yard,  and  thereby  make  25 
per  cent.,  what  was  its  first  cost  ?  Ans.  $  4  per  yard. 

operation.  Since  the  gain  is  25  per  cent,  of 

$5.00  -T-  1.25  =  $4,  Ans.     the  cost,  the   selling  price,  $5,  is 

equal  to  the  cost  increased  by  25 
per  cent,  of  the  cost,  or  1.25  of  the  cost.  Therefore,  the  cost  must 
be  as  many  dollars  as  5  contains  times  1.25. 

Rule.  —  Divide  the  selling  price  by  1  increased  by  the  gain  per  cent., 
or  by  1  decreased  by  the  loss  per  cent.,  expressed  decimally,  and  the  quo- 
tient will  be  the  cost. 

Examples. 

2.  If  I  dispose  of  cloth  at  $  4  per  yard,  and  by  so  doing  lose 
20  per  cent.,  required  the  prime  cost  of  the  goods  ?  Ans.  $  5. 

3.  Sold  10  barrels  of  flour  for  $  96,  and  made  20  per  cent. 
What  was  the  prime  cost  a  barrel  ?  Ans.  $  8. 

4.  If  27^cwt.  of  sugar  be  sold  at  $  12.50  per  cwt.,  and  there 
is  gained  17  per  cent.,  what  was  the  prime  cost  per  cwt.  ? 

5.  Sold  wood  at  $  6.1 2J  per  cord,  and  by  so  doing  lost  12^ 
per  cent,  a  cord.     What  was  the  original  cost  per  cord  ? 

Ans.  $  7.00. 


PROFIT   AND   LOSS.  305 

414.     To  find  the  gain  or  loss  per  cent,  when  the  cost  and 
selling  price  are  given. 

Ex.  1.     If  I  buy  cloth  at  $  4,  and  sell  it  at  $5  per  yard, 
what  per  cent,  do  I  gain  ?  Ans.  25  per  cent. 


OPERATION. 


$5  —  $4  =  $1;  1.00  -r-  4  =  .25,  or  25  per  cent.,  Ans. 

Since  the  difference  between  the  selling  price  and  the  prime  cost 
is  $  1,  the  gain  is  ^  of  the  cost,  or,  expressed  decimally,  is  .25  of  the 
cost,  or  25  per  cent,  of  it. 

Rule.  — Divide  the  number  denoting  the  gain  or  loss  by  that  denot- 
ing the  prime  cost,  and  the  quotient,  expressed  decimally,  will  be  the 
gain  or  loss  per  cent,  required. 

Examples. 

2.  Bought  cloth  at  $7  per  yard,  and  sold  it  at  $  6.12£. 
What  per  cent,  did  I  lose?  Ans.  12^  per  cent. 

3.  Bought  a  chaise  for  $ 200,  and  sold  it  for  $225.  What 
per  cent,  did  I  gain  ?  Ans.  12  J-  per  cent. 

4.  I  sell  a  house  that  cost  me  $  2500  at  an  advance  of 
$  500.     What  per  cent,  do  I  gain  ? 

5.  Bought  24  yards  of  cloth  for  $  64.8  64ff,  and  sold  it  at 
$  2.50  per  yard.     What  per  cent,  is  the  loss  ? 

;  Ans.  7£  per  cent. 

415.  The  selling  price  of  goods  and  the  rate  per  cent,  of 
gain  or  loss  being  given,  to  find  what  the  gain  or  loss  per  cent, 
would  be,  if  sold  at  another  price. 

Ex.  1.  If  I  sell  cloth  at  $  5  per  yard,  and  thereby  gain  25 
per  cent.,  what  would  have  been  my  gain  if  I  had  sold  it  at  $  7 
per  yard  ?  Ans.  75  per  cent. 

operation.  -\ye  find  the  prime 

$  5.00  ~  1.25  =  $4;  $7  —  $4=$  3.     cost  of  the  cloth  per 

3.00  — -  4  =  .75,  or  75  per  cent.,   Ans.       yard,  when  sold  at 

$5,  by  Art.  413,  to 
be  $  4.  We  then  find  what  would  have  been  the  gain  per  cent,  on 
the  cost,  if  it  had  been  sold  at  $  7  per  yard,  by  Art.  414,  and  obtain 
75  per  cent,  as  the  answer. 

Rule.  —  Find  the  prime  cost  (Art.  413),  and  then  the  gain  or  loss 
per  cent,  on  this  cost  at  the  proposed  selling  price. 
26* 


306  MISCELLANEOUS   EXAMPLES. 

Examples. 

2.  If  I  purchase  cloth  at  $  7  per  yard,  and  thereby  gain  75 
per  cent.,  do  I  gain  or  lose  if  I  sell  the  same  at  $  3  per  yard  ? 

Ans.  Lose  25  per  cent. 

3.  Bought  wheat  at  $  1.25  per  bushel,  and  lost  15  per  cent. 
What  per  cent,  should  I  have  gained  had  I  sold  it  for  $1.G47T1T 
per  bushel  ?  Ans.  12  per  cent. 

4.  Sold  wheat  at  $  1.647Ty  per  bushel,  and  gained  12  per 
cent.  What  per  cent,  should  I  have  lost  had  I  sold  it  for  $  1.25 
per  bushel  ? 

5.  I  sold  a  horse  for  $  75,  and  by  so  doing  I  lost  25  per 
cent. ;  whereas,  I  ought  to  have  gained  30  per  cent.  How 
much  was  he  sold  under  his  real  value  ?  Ans.  $  55.00. 

6.  When  tea,  sold  at  a  loss  of  25  per  cent.,  brings  $  1.25  per 
pound,  what  would  be  the  gain  or  loss  should  it  bring  $  1.40 
per  pound?  Ans.  16  per  cent.  loss. 


MISCELLANEOUS   EXAMPLES. 

1.  Bought  stocks  whose  par  value  is  $  100  per  share,  and 
sold  the  same  at  a  premium  of  15  per  cent,  and  thereby  gained 
$  120.     How  many  shares  were  there?  Ans.  8  shares. 

2.  Mining  stock  of  a  par  value  of  $  250  per  share  was 
bought  at  9  per,  cent,  premium,  and  afterwards  sold  at  a  loss  of 
$  25  per  share  on  the  price  paid.  At  what  rate  per  cent,  was 
it  sold  ?  Ans.  1  per  cent,  discount. 

3.  How  much  greater  income  may  be  realized  yearly  from 
$  19200  invested  in  7  per  cent,  stocks  purchased  at  9G  per 
cent.,  than  from  the  same  amount  in  5  per  cent,  stocks  pur- 
chased at  80  per  cent.  ? 

4.  The  par  value  of  the  shares  in  a  certain  manufacturing 
company  is  $  250  each,  and  the  regular  yearly  dividends  are 
$  15  a  share.  At  what  price  should  the  shares  be  bought  that 
the  investment  may  pay  10  per  cent,  interest? 

Ans.  $  150  each. 


MISCELLANEOUS   EXAMPLES.  807 

5.  A  broker  charges  me  1£  per  cent,  brokerage  for  purchas- 
ing some  uncurrent  bank-bills  at  20  per  cent,  discount.  Of 
these  bills  four  of  $  50  each  become  worthless,  but  the  remain- 
der I  dispose  of  at  par,  and  thus  make  by  the  operation  $  364. 
What  was  the  amount  of  the  bills  ?  Ans.  $  3000. 

6.  A  commission  merchant  sold  an  assignment  of  cotton  for 
$  5640,  and  after  deducting  his  commission,  and  $  76.50  for 
freight,  storage,  &c,  remitted  to  his  assignee  $  5422.50  as  the 
net  proceeds.     What  per  cent,  was  his  commission  ? 

Ans.  2J  per  cent. 

7.  A  horse  which  I  bought  for  30  per  cent,  less  than  his  real 
worth,  having  become  injured,  I  sold  him  for  25  per  cent,  less 
than  what  he  cost,  and  thereby  lost  $  55  of  his  original  value. 
What  did  I  get  for  the  horse  ?  Ans.  $  60.78||. 

8.  A  watch,  which  cost  me  $  30,  I  have  sold  for  $  35,  on  a 
credit  of  8  months.     What  did  I  gain  by  the  bargain  ? 

Ans.  $3.653|J. 

9.  Bought  a  hogshead  of  molasses  for  $  112 ;  but  15  gallons 
having  leaked  out,  the  remainder  was  sold  at  $  2.21 6§  a  gallon. 
What  per  cent,  is  the  loss  ?  Ans.  5  per  cent. 

10.  A  hogshead  of  molasses  was  bought  for  a  certain  sum ; 
but  15  gallons  having  leaked  out,  the  remainder  was  sold  at 
$  2.21 6§  a  gallon,  and  thereby  the  loss  was  5  per  cent,  on  the 
cost.     What  was  the  cost  ?  Ans.  $  112. 

11.  Bought  50  barrels  of  flour  at  $9.00  per  barrel,  but  a 
part  of  it  having  been  damaged,  half  of  it  was  sold  at  a  loss  of 
10  per  cent.,  and  the  remainder  at  $  9.50  per  barrel.  How 
much  was  lost  by  the  operation  ?  Ans.  $  10. 

12.  My  agent  purchases  flour  for  me,  which,  with  his  com- 
mission at  2£  per  cent,  cost  $  6135,  and  I  dispose  of  it  at  20 
per  cent,  advance  upon  the  price  he  paid.  What  did  I  make 
by  the  operation,  allowing  freight  and  storage  to  have  been 
$31.63? 

13.  I'  have  remitted  to  my  correspondent  a  certain  sum  of 
money,  which  he  invested  in  iron ;  and  having  reserved  to  him- 
self 2£  per  cent,  on  the  purchase,  which  amounted  to  $  90,  he 
purchased  the  iron  at  $  95  per  ton.  Required  the  sum  re- 
mitted, and  the  quantity  of  iron  purchased. 

.        \  Sum  remitted,  $  3690. 

"  |  Iron  purchased,  37T.  17cwt.  3qr.  14T°alb. 


308  PARTNERSHIP,    OR   COMPANY   BUSINESS. 


PARTNERSHIP,  OR  COMPANY  BUSINESS. 

416*  Partnership  is  the  association  of  two  or  more  per- 
sons in  business,  with  an  agreement  to  share  the  profits  and 
losses  in  proportion  to  the  amount  of  the  capital  stock,  or  the 
value  of  the  labor  and  experience  of  each. 

Partners  are  the  persons  associated  in  business. 

Company,  or  Firm,  is  the  general  name  of  the  business  asso- 
ciation. 

Capital,  or  Joint  Stock,  is  the  money  or  property  invested  in 
the  company  or  firm. 

Dividend  is  the  profit  or  loss  on  the  shares  of  the  capital,  or 
joint  stock. 

Note.  —  Partnership  is  an  application  of  the  principle  of  distributive  or  par- 
titive proportion  (Art.  339). 

41 7i  To  find  each  partner's  share  of  the  profit  or  loss,  when 
there  is  no  regard  to  time. 

Ex.  1.  Three  men,  A,  B,  and  C,  enter  into  partnership  for 
two  years,  with  a  capital  of  $  1080.  A  puts  in  $  240,  B  $  360, 
and  C  $  480.  They  gain  $  54.  What  is  each  man's  part  of 
the  gain  ? 

OPERATION. 

A's  stock,  $  2  4  0     $  2  4  0  X  .0  5  —  $  1  2,  A's  gain. 

B's      "         3  6  0        3  6  0  X  .0  5  =     1  8,  B's  gain. 

C's      "        48  0        480  X  -05  sfc     24,  W  gain. 

Entire  stock,  $  1  0  8  0  Proof,  $  5  4,  entire  gain. 

§54.00  -T-  1080  =  $0.05,  gain  on  $1. 

Since  the  entire  stock  is  $  1080  and  the  entire  gain  $  54,  the  gain 
on  every  $  1  of  stock  will  be  as  many  dollars  as  54  contains  times 
1080,  or  $  0.05  on  every  $  1  of  stock.  Then,  each  man's  stock  mul- 
tiplied by  .05  gives  his  part  of  the  entire  gain.  The  same  result  also 
may  be  obtained,  as  follows, 

BY    PROPORTION. 

$1080:  $240  ::  $54:  $  1  2,  A's  gain. 
$1080:8360  ::  $  5  4  :  $  1  8,  B's  gain. 
$1080  :  $480  ::  $54:  $2  4,  (7s  gain. 

Proof,  $  5  4,  entire  gain. 
Rulk.  —  The  entire  gain  or  loss,  divided  by  the  number  denoting  the 


PARTNERSHIP,   OR   COMPANY  BUSINESS.  309 

entire  stock,  icill  give  the  gain  or  loss  on  each  dollar  of  stock ;  and  each 
partner's  stock,  multiplied  by  the  number  denoting  the  gain  on  $  1,  icill 
give  his  share  of  the  entire  gain  or  loss.     Or, 

As  the  whole  stock  is  to  each  partner's  stock,  so  is  the  whole  gain  or 
loss  to  each  partner 's  gain  or  loss. 

Examples. 

2.  Jones,  Weston,  and  Sprague  traded  in  company,  with  a 
capital  of  $  10000  ;  Jones  put  in  $  3000,  Weston  $  2000,  and 
Sprague  $  5000  ;  they  gained  $  4000.  What  was  each  man's 
part  of  the  gain  ? 

Ans.  Jones's  part,  $  1200  ;  Weston's  part,  $  800  ;  Sprague's 
part,  $  2000. 

3.  Two  merchants,  C  and  D,  engaged  in  trade;  C  put  in 
$  6780,  and  D  put  in  $  12000  ;  they  gain  $  2000.  What  was 
each  man's  part  ? 

Ans.  C's  part,  722.044;  D's  part,  $  1276.955. 

4.  Harvey,  Blake,  and  Horsford  entered  into  partnership 
with  a  capital  of  $11000,  of  which  Harvey  put  in  $2500, 
Blake  $  3000,  and  Horsford  $  5500 ;  they  lost  by  trading  5 
per  cent,  on  their  capital.  What  was  each  partner's  share  of 
the  loss? 

5.  Elliott,  May  hew,  and  Griswold  engaged  together  in  a 
speculation.  Of  the  money  employed  Elliott  furnished  $  500 ; 
Mayhew  $  350,  and  Griswold  a  cart  and  two  horses ;  they 
gained  $332.50,  of  which  Griswold's  part  was  $120.  What 
were  Elliott's  and  Mayhew's  parts  of  the  gain,  and  what  was 
the  value  of  Griswold's  part  of  the  stock  ? 

Ans.  Elliott's  gain,  $125;  Mayhew's  gain,  $87.50;  Gris- 
wold's stock,  $  480. 

6.  A,  B,  and  C  traded  in  company  ;  A  put  in  $  5000,  B  put 
in  $  6500,  and  C  put  in  $  7500  ;  they  gain  40  per  cent,  on  their 
capital,  but  receive  the  whole  amount  of  their  gains  in  bills,  for 
which  they  are  obliged  to  allow  a  discount  of  10  per  cent. 
How  much  was  each  man's  net  gain  ? 

Ans.  A's  gain,  $1800;  B's  gain,  $2340;  C's  gain,  $2700. 

7.  A,  B,  C,  and  D  are  in  partnership,  with  a  joint  capital  of 
$  40,000  ;  on  dividing  their  profits,  it  is  found  that  A's  share  is 
$2000,  B's  share  $4500,  C's  share  $2500,  and  D's  share 
$  1500.     What  was  each  partner's  stock  ? 


310  PARTNERSHIP,   OR   COMPANY   BUSINESS. 

8.  A,  B,  and  C  were  associated  in  trade ;  A's  part  of  the  gen- 
eral stock  was  $  2000,  B's  part  $  3000,  and  C's  part  $  7500. 
On  dividing  the  profits,  it  was  found  that  A's  and  B's  gain  to- 
gether amounted  to  $  1000,  which  was  $  500  less  than  C's 
gain.     What  was  the  gain  of  each  ? 

Ans.  A's  gain,  $  400 ;  B's  gain,  $  600  ;  C's  gain,  $'1500. 

9.  A,  B,  and  C  own  a  ship  together,  which  cost  them 
$  30000 ;  of  which  A  paid  an  unknown  sum,  B  paid  1£  as 
much,  and  C  paid  1^  as  much.  The  profits  were  25  per  cent, 
of  the  cost  of  the  ship.     What  was  each  man's  gain  ? 

Ans.  A's  gain,  $  2000 ;  B's  gain,  $  3000  ;  C's  gain,  $  2500. 

10.  Walker,  Edwards,  and  Armstrong  are  partners,  whose 
respective  shares  of  joint  stock  are  as  the  fractions  •£,  £,  and  -J. 
They  gain  $  50000.  If,  on  dividing  the  profits,  Armstrong  re- 
linquishes his  part  of  the  gain,  how  much  will  each  of  the  others 
receive  ? 

Ans.  Walker,  $  28571.42§§f  ;  Edwards,  $  21428.57^. 

418.  To  find  each  partner's  share  of  the  profit  or  loss,  when 
the  stock  is  employed  for  different  periods  of  time. 

Ex.  1.  A  and  B  are  associated  in  trade.  A  has  furnished 
of  the  joint  stock  $  420  for  5  months,  and  B  has  furnished 
$  350  for  8  months ;  their  net  profits  are  $  84.  What  is  each 
man's  share  of  the  gain?       Ans.  A's  share,  $36  ;  B's,  $48. 

OPERATION. 

$420  X  5  =  $2100    $2100  X  .01$  =  $36,  A's  gain. 
$350XS  =  $2800    $  2  8  0  0  X  .0  1  f  =  $  4  8,  B's  gain. 

$4900  Proof,  $84,  entire" 

It  is  evident  that  $420  for  5  months  is  the  same  as  $  420  x  5  = 
$  2100  for  1  month,  since  $  2100  would  gain  as  much  in  1  month  as 
$ 420  in  5  months;  and  $  350  for  8  months  is  the  same  as  $  350  X  8 
=  $  2800,  for  1  month.  The  question  then  is  the  same  as  if  A  had 
furnished  $  2100  and  B  $  2800  for  equal  times.  Then,  if  $  2100  4- 
$  2800  ==  $  4900,  gain  $84,  $  1  will  gain  ^^  of  $  84  =  $  O.Olf , 
and  $2100  X  -01f  =  $36,  A's  gain;  $2800  X  .01f  =  $48,  1^'s 
gain.     The  same  results  may  be  obtained,  as  follows, 

BY    PROPORTION'. 

$4900:$2100::$84:$3  6,  A's  gain. 
$4  9  00:  $2  800::  $8  4:  $48,  B's  gain. 

Proof,  $  8  4,  entire  gain. 


PARTNERSHIP,   OR   COMPANY   BUSINESS.  311 

Rule.  —  Multiply  each  partner's  stock  by  the  time  it  was  in  trade, 
and  divide  the  entire  gain  or  loss  by  the  sum  of  the  several  products  ;  by 
the  quotient  multiply  the  product  of  each  partner's  stock  and  time,  and 
the  result  will  be.  his  share  of  the  gain  or  loss.     Or, 

Multiply  each  partner's  stock  by  the  time  it  was  in  trade  ;  then,  as  the 
sum  of  these  products  is  to  each  product,  so  is  the  whole  gain  or  loss  to 
each  partner's  gain  or  loss. 

Examples. 

2.  Goodwin  commenced  business  January  1,  with  a  capital 
of  $  3200  ;  May  1,  he  took  Blunt  into  partnership,  with  a  capi- 
tal of  $  4200 ;  and  at  the  end  of  the  year  they  had  gained 
$  240.     What  was  each  partner's  share  of  the  gain  ? 

Ans.  Goodwin's  gain,  $  128 ;  Blunt's  gain,  $  112. 

3.  Three  men  hire  a  pasture  in  common,  for  which  they  are 
to  pay  $  26.40.  A  put  in  24  oxen  for  8  weeks,  B  put  in  18 
oxen  for  12  weeks,  and  C  put  in  12  oxen  for  10  weeks.  What 
ought  each  to  pay  ? 

4.  Barclay,  Hickman,  and  Oliver  are  partners.  Barclay 
furnishes  of  the  capital  $  300  for  5  months,  Hickman  $  400  for 
8  months,  and  Oliver  $500  for  3  months;  they  gain  $200. 
After  paying  $  50  for  advertising  and  $  50  for  agency,  what 
will  be  each  partner's  share  of  the  net  profits  ? 

Ans.  Barclay's   share,   $24.19£|;    Hickman's,   $51.61^; 
Oliver's,  $24.19^. 

5.  A,  B,  and  C  engaged  in  partnership,  with  a  joint  capital 
of  $  1000,  A  putting  in  stock  for  7  months,  B  for  8  months,  and 
C  for  12  months.  Of  the  profits  A's  part  was  $  21 ;  B's,  $  40  ; 
and  C's,  $  24.     Required  the  capital  each  put  in. 

Ans.  A,  $  300 ;  B,  $  500 ;  C,  $  200. 

6.  White  and  Daniels  traded  in  company,  with  a  joint  stock 
of  $  6300.  White's  money  having  been  employed  12  months, 
and  Daniel's  8  months,  on  dividing  profits,  each  had  gained 
exactly  the  same  sum.  How  much  of  the  capital  did  each 
furnish  ? 

7.  Three  men  engage  in  partnership  for  20  months.  A  at 
first  put  into  the  firm  $  4000,  at  the  end  of  four  months  he 
put  in  $  500  more,  and  at  the  end  of  16  months  he  took  out 
$  1000;  B  at  first  put  in  $3000,  at  the  end  of  10  months 
he  took  out  $  1500,  and  at  the  end  of  14  months  he  put  in 


.312  PARTNERSHIP,   OR   COMPANY   BUSINESS. 

$3000;  C  at  first  put  in  $2000,  at  the  end  of  G  months 
he  put  in  $  2000  more,  at  the  end  of  14  months  he  put  in 
$2000  more,  and  at  the  end  of  16  months  he  took  out  $  1500 ; 
they  had  gained  by  trade  $  4420.  What  is  each  man's  share 
of  the  gain  ? 

Ans.  A's  gain,  $  1680  ;  B's  gain,  $  1260 ;  C's  gain,  $  1480. 

8.  Grover  and  Thorndike  are  associated  in  trade,  Grover 
contributing  of  the  capital  $  12000,  and  Thorndike  $  18000. 
At  the  end  of  6  months  they  reduce  the  joint  stock  $  5000,  by 
each  withdrawing  an  equal  sum.  3  months  afterwards  Thorn- 
dike withdraws  $6000,  and  Grover  $1000.  The  business 
proving  a  losing  one,  they  dissolve  copartnership  at  the  end  of 
the  year.  Required  what  part  of  the  stock  then  remaining, 
which  was  only  $  15000,  belonged  to  each  of  the  partners  ? 

Ans.  To  Grover,  $6116^;  to  Thorndike,  $88837W 

9.  John  Jones,  Samuel  Eaton,  and  Joseph  Brown  formed  a 
partnership,  under  the  firm  of  Jones,  Eaton,  &  Co.,  with  a  capi- 
tal of  $  10,000  ;  of  which  Jones  put  in  $  4000,  Eaton  put  in 
$  3500,  and  Brown  put  in  $  2500  ;  but  at  the  end  of  6  months 
Jones  withdrew  $  2000  of  his  stock,  and  at  the  end  of  8  months 
Eaton  withdrew  $  1500  from  the  firm ;  but  at  the  end  of  10 
months  Brown  added  $  2000  to  his  stock.  At  the  end  of  2 
years  they  found  their  gains  to  be  $  1041.80.  What  was  the 
share  of  each  man  ? 

Ans.  Jones's  gain,  $  300.51|§ ;  Eaton's  gain,    $300.51|§; 
Brown's  gain,  $440.76T2^. 

10.  James  Bradshaw  commenced  trade,  January  1,  1856, 
with  a  capital  of  $  10000,  and  after  some  time  formed  a  part- 
nership with  John  Parkman,  who  contributed  to  the  joint  stock 
$  2800.  In  course  of  time  they  admitted  into  the  firm  Joseph 
Delano,  with  a  stock  of  $3600.  On  making  a  settlement, 
January  1,  1857,  it  was  found  that  Bradshaw  had  gained 
$  2250  ;  Parkman,  $  420 ;  and  Delano,  $  405.  How  long  had 
Parkman's  and  Delano's  money  been  employed  in  trade,  and 
what  rate  of  interest  per  annum  had  each  of  the  partners 
gained  on  their  stock  ? 

Ans.  Parkman's,  8  months ;  Delano's,  6  months.    Gained  22£ 
per  cent,  interest. 


BANKRUPTCY.  313 


BANKRUPTCY. 

419t  Bankruptcy  refers  to  business  failures  and  inability 
to  meet  pecuniary  liabilities. 

A  bankrupt,  or  insolvent,  is  one  who  fails  or  becomes  unable 
to  pay  his  debts. 

An  assignee  is  a  person  selected  to  take  charge  of  the  prop- 
erty and  effects  of  a  bankrupt,  to  convert  the  same  into  cash, 
and,  after  deducting  the  necessary  expenses  of  settling,  to  di- 
vide the  net  proceeds,  as  the  law  requires,  among  the  creditors. 

The  distribution  is  generally  made  pro  rata,  each  creditor 
receiving  according  to  his  respective  demand,  or  just  claim. 

420.  To  find  each  creditor's  dividend  or  share  of  the  net 
proceeds  of  an  insolvent  estate. 

Ex.  1.  A  bankrupt  owes  to  A  $500,  to  B  $1200,  and  to 
C  $  4300  ;  and  the  net  cash  proceeds  of  his  estate  amount  to 
only  $  1500.  How  much  does  he  pay  on  $  1,  and  what  divi- 
dend does  each  creditor  receive  ? 

Ans.  25  cents  on  a  dollar.  A  receives  $  125 ;  B,  $  300 ; 
and  C,  $  1075. 

OPERATION. 

A's  claim,  $  500  $  500  X  -25  =  $  125,  A's  share. 
B's  "  $  1200  $  1200  X  .25  =  $  300,  B's  share. 
C's      "     $  4300     $  4300  X  .25  =  $  1075,  C's  share. 

Entire  claims,  $  6000  Proof,  $  1500,  entire  proc'ds. 

$  1500.00  ~  6000  —  $  0.25,  or  25  cents  on  $  1. 

Rule.  — Divide  the  net  proceeds  of  the  insolvent  estate  by  the  num- 
ber denoting  the  total  amount  of  its  indebtedness,  to  find  the  sum  it  pays 
on  each  dollar  of  the  debts. 

Multiply  each  man's  claim  by  the  sum  the  estate  pays  on  a  dollar,  to 
obtain  each  man's  dividend. 

Examples. 

2.  Clarke,  Soule,  &  Co.  have  failed.  Their  liabilities  are 
$  63500,  their  assets  have  a  cash  value  of  $  52384,  and  the 
expenses  of  settling  are  $  1584.  How  much  can  they  pay  on 
a  dollar,  and  what  dividend  should  John  Dayton  receive,  whose 
claim  is  $  8361.55  ?  Ans.  80  cents  on  a  dollar ;  $  6689.24. 
N      27 


314  TAXES. 

3.  A  merchant  failing  in  trade  owes  A  $  GOO,  B  $760,  C 
$840,  and  D  $800.  The  net  proceeds  of  his  effects  are 
$  2275.     What  dividend  does  each  of  his  creditors  receive  ? 

Ans.  A,  $455  ;  B,  $576.33£;  C,  $  637  ;  D,  $  GOG.GGf. 

4.  J.  Bonney  owes  A  $  400,  B  $  300,  and  C  $  1000.  His 
effects  are  worth  $  600.  What  sum  can  he  pay  each  of  his 
creditors  ?    • 

5.  A  manufacturing  company  becomes  insolvent.  Its  indebt- 
edness amounts  to  $300000.  Its  assets  consist  of  factory 
buildings  and  machinery  worth  $  180000,  stock  worth  $40000, 
and  bills  receivable  good  for  $  12875.  The  charges  of  the 
court  of  insolvency  and  the  assignee  will  amount  to  3£  per 
cent,  on  the  amount  distributed  to  the  creditors.  How  much 
will  the  company  pay  on  a  dollar,  and  what  will  be  Amos 
Henderson's  dividend  on  a  claim  of  $  1360.60  ? 

Ans.  75  cents  on  a  dollar;  $  1020.45. 


TAXES. 


421.  A  tax  is  a  sum  of  money  assessed  on  individuals  by 
government,  corporations,  societies,  districts,  &c. 

Taxes  for  government  purposes  are  imposed  on  property, 
and  in  most  States  on  persons. 

A  poll  or  capitation  tax  is  one  without  regard  to  property, 
on  the  person  of  each  male  citizen  liable  by  law  to  assessment. 
A  person  thus  liable  is  termed  a  poll. 

422.  Immovable  property,  such  as  lands,  mills,  houses, 
&c.,  is  called  real  estate.  All  other  property,  such  as  money, 
notes,  mortgages,  cattle,  shipping,  furniture,  &c,  is  called  per- 
sonal  property. 

423.  In?  assessing  taxes,  it  is  necessary  to  have  an  in- 
ventory of  taxable  property ;  and,  if  a  levy  on  the  polls  is 
to  be  included,  there  should  be  also  a  complete  list  of  taxable 
polls. 


TAXES.  315 

424.  The  method  of  assessing  taxes,  though  not  precisely 
the  same  in  all  the  States,  is  yet  in  most  of  them  virtually  the 
same. 

In  some  of  the  States,  however,  the  public  schools  are  sup- 
ported in  whole  or  in  part  by  rate  bills  ;  that  is,  the  expenses  of 
the  schools,  in  whole  or  in  part,  are  apportioned  among  the  in- 
habitants patronizing  the  same,  according  to  the  number  of  days 
of  each  pupil's  attendance. 

Note.  —  In  Massachusetts  one  sixth  part  of  the  whole  sum  to  be  raised  is 
assessed  upon  the  polls ;  provided  the  whole  poll  tax  assessed  in  any  one  year 
upon  any  individual  for  town  and  county  purposes,  except  highway  taxes, 
shall  not  exceed  one  dollar  and  fifty  cents. 

425.  To  assess  a  state,  county,  town,  or  other  tax. 

Ex.  1.  The  inhabitants  of  a  certain  town  are  to  be  taxed 
$4109.  The  real  estate  of  the  town  is  valued  at  $493000, 
and  the  personal  property  at  $  177000.  There  are  506  polls, 
each  of  which  is  to  be  taxed  $  1.50.  What  is  the  tax  on  each 
dollar  of  property  ?  What  is  J.  B.  Tewksbury's  tax,  whose  real 
estate  is  valued  at  $  3700,  and  his  personal  at  $  2300,  he  pay- 
ing for  2  polls  ? 

OPERATION. 

$  1.50  X  506  =  $759,  amount  assessed  on  the  polls. 

$  493000  +  $177000  =  $  670000,  amount  of  taxable  property. 

$  4109  —  $  759  =;  $  3350,  amount  to  be  assessed  on  property. 

$  3350  -r-  670000  —  $  .005,  tax  to  be  assessed  on  each  dollar. 

$  3700  -f  $  2300  =  $  6000,  Tewksbury's  taxable  property. 

$  6000  X  .005  =  $  30,  Tewksbury's  property  tax. 

$1.50  X  2  =  $3,  Tewksbury's  poll  tax. 

$  30  -\-  3  =  $  33,  amount  of  Tewksbury's  tax. 

Rule.  —  Multiply  the  number  of  taxable  polls  by  the  tax  on  each 
poll ;  and  the  product,  subtracted  from  the  whole  sum  to  be  raised,  will 
give  the  sum  to  be  raised  on  the  property. 

The  sum  to  be  raised  on  property,  divided  by  the  whole  taxable  prop- 
erty, will  give  the  sum  to  be  paid  on  each  dollar  of  property  taxed. 

Each  man's  taxable  property,  multiplied  by  the  number  denoting  the 
sum  to  be  paid  on  $1,  with  his  poll  tax  added  to  the  product,  will  give 
the  amount  of  his  tax. 

Note.  —  The  operation  of  assessing  taxes  may  be  facilitated  by  finding  the 
tax  on  $  2,  $  3,  &c.  at  the  rate  of  taxation  on  $  1,  before  making  the  assess- 
ment on  the  inhabitants  of  the  town,  &c,  and  arranging  the  numbers  as  in  the 
following 


316 


TAXES. 
Table. 


Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

$1 

$0,005 

$20 

$0.10 

$300 

$1.50 

$4,000 

$20.00 

2 

0.010 

30 

0.15 

400 

2.00 

5,000 

25.00 

3 

0.015 

40 

0.20 

500 

2.50 

6,000 

30.00 

4 

0.020 

50 

0.25 

600 

3.00 

7,000 

35.00 

5 

0.025 

60 

0.30 

700 

3.50 

8,000 

40.00 

G 

0.030 

70 

0.35 

800 

4.00 

9,000 

45.00 

7 

0.035 

80 

0.40 

900 

4.50 

10.000 

50.00 

8 

0.040 

90 

0.45 

1,000 

5.00 

20,000 

100.00 

9 

0.045 

100 

0.50 

2,000 

10.00 

30,000 

150.00 

10 

0.050 

200 

1.00 

3,000 

15.00 

40,000 

200.00 

Examples. 

2.  What  is  Samuel  Forster's  tax,  by  the  above  table,  whose 
property,  real  and  personal,  is  valued  at  $1310,  and  who  pays 
for  7  polls  a  tax  of  $  1.50  each  ?  Ans.  $  17.05. 

3.  What  is  the  tax  of  a  non-resident,  having  property  in  the 
same  town,  worth  $  115.35  ? 

4.  A  tax  of  $14018.90  is  to  be  levied  on  a  certain  city. 
The  property  is  valued  at  $  3506300 ;  and  there  are  3500 
polls,  each  of  which  is  taxed  $  1  each.  What  is  the  assessment 
Oil  a  dollar,  and  what  is  A's  tax,  who  has  property  worth  $  29010, 
and  pays  for  2  polls  ?    Ans.  $  .003  on  $  1 ;  A's  tax,  $  89.03. 

426.  To  find  what  sum  must  be  assessed  to  yield  a  given 
net  amount. 

Ex.  1.  What  sum  must  be  assessed  to  yield  a  net  amount 
of  $  6240,  the  collectors  receiving  2£  per  cent,  commission  for 
collecting?  Ans.  $6400. 


Since   the   commission   is    1\  per 
cent.,  the  net  value  of  each  dollar 


OPERATION'. 

$  6240  ~  .975  =£  $  6400. 

of  assessment  will  be  $0,975,  and 
the  amount  to  be  assessed  to  net  $  6240  will  therefore  be  as  many 
dollars  as  6240  contains  times  .975.     Hence, 

The  given  net  amount  divided  by  1  decreased  by  the  rate  of  commis- 
sion, expressed  decimally,  roill  give  the  amount  to  be  assessed. 

Examples. 

2.  What  sum  must  be  assessed  to  net  $  10450,  allowing  the 
collector  receives  5  per  cent,  for  collecting  ?     Ans.  $  11000. 


TAXES.  317 

3.  Allowing  the  net  sum  to  be  raised  by  a  society  to  be 
$  9700,  and  the  allowance  for  collection  to  be  3  per  cent.,  what 
is  the  gross  amount  to  be  assessed,  and  how  much  will  be  the 
cost  of  collection  ? 

Ans.   Assessment,  $  10000  ;  cost  of  collection,  $  300. 

4.  The  taxable  property  in  a  certain  district  containing  450 
polls  is  $  756000.  It  is  proposed  to  raise  $  18000  for  building 
a  union  school-house.  If  the  poll  tax  be  limited  to  $  1.50  a 
poll,  and  the  cost  of  collecting  be  3  per  cent.,  what  will  be  the 
required  assessment  on  a  dollar,  and  how  much  will  be  A's  tax, 
who  pays  for  3  polls,  and  has  property  to  the  amount  of 
$  15600.  Ans.  Tax  on  $  1,  $  .024 ;  A's  tax,  $  378.90. 

427  •  To  apportion  school  expenses  according  to  each  pupil's 
attendance,  or  to  make  out  rate  bills. 

Ex.  1.     The  expenses  of  a  certain  district  school  are  $  464, 

and  the  aggregate  of  attendance  9280  days.    What  is  A's  tax, 

who  has  sent  pupils  amounting  to  157  days  ? 

operation  Since  9280  days'  attendance  cost 

$464.00  -r-  9280  =  $0.05    $464'  x  da/'?,  attenda"c?  will  cost 

as    many  dollars  as  464   contains 
$0.05  X  157  sa  $7.85.        times  9280,  and  157  days  will  cost 

157  times  as  much  as  1  day.  Hence, 

The  wliole  expenses  of  the  school,  less  the  public  money,  if  any,  divided 
by  the  number  of  days'  attendance,  will  give  the  rate  for  each  day's  at- 
tendance. 

The  rate  for  each  day,  multiplied  by  the  number  of  days  of  attendance 
of  each  patron's  pujnls,  will  give  the  amount  of  his  rate  bill. 

Examples. 

2.  If  a  district  expends  for  a  teacher's  salary  $  500,  for  his 
board  $  150,  for  repairs  of  school-house  $  30  ;  the  public  money 
being  $350,  and  the  whole  number  of  days  of  attendance  5500, 
what  is  the  rate  per  day,  and  what  is  A's  bill,  who  sends  2  pu- 
pils 60  days  each,  and  1  pupil  30  days  ? 

Ans.   Rate  per  day,  $  0.06 ;  A's  bill,  $  9.00. 

3.  In  a  certain  district  the  teacher's  wages  amounted  to 
$  150,  the  fuel  cost  $  18.50,  the  public  money  received  was 
$  63.50,  and  the  number  of  days'  attendance  was  3000.  What 
was  A's  rate  bill,  whose  pupils  attended  121  days  ;  and  B's, 
whose  pupils  attended  173  days  ? 

27* 


318  GENERAL  AVERAGE. 


GENERAL  AVERAGE. 


428.  General  Average  signifies  a  contribution  ratably 
made  to  a  general  loss  by  the  three  great  mercantile  interests, 
vessel,  freight,  and  cargo,  —  when,  on  account  of  a  common  peril, 
one  or  more  of  these  has  been  voluntarily  sacrificed,  in  whole 
or  in  part,  to  effect  the  preservation  of  the  rest. 

Jettison  is  the  term  applied  to  the  part  of  the  vessel  or  cargo 
cast  overboard,  or  otherwise  sacrificed,  to  save  the  remainder. 

429.  Particular  average  means  an  average  to  which  con- 
tribute only  "  some  parts  of  the  property  when  severed  from  the 
whole ;  as  in  the  case  of  a  boat-load  saved  from  peril  by  a  sacri- 
fice, to  which  only  the  boat  and  its  contents  contribute.',  Prac- 
tically, it  hardly  needs  to  be  discriminated  from  general  average, 
since  it  is  adjusted  upon  its  contributory  interest  by  the  applica- 
tion of  the  same  principles. 

430.  To  constitute  a  valid  claim  to  general  average,  there 
are  three  essentials,  all  of  which  must  be  present:  1.  A  com- 
mon peril  impending  at  the  time  ;  2.  A  voluntary  and  premedi- 
tated sacrifice  of  some  property  for  the  purpose  of  saving  other 
property ;  3.  The  success  of  the  endeavor.  (Parsons  on  Mer- 
cantile Law.) 

431.  In  adjusting  a  general  average,  the  property  sacrificed, 
as  well  as  that  saved,  is  regarded  as  contributory  to  the  general 
loss.  The  entire  value  of  the  freight,  however,  is  not  contrib- 
utory ;  i  of  its  value  in  New  York,  and  ^  of  its  value  in  other 
parts  of  the  United  States,  being  reserved  for  seamen's  wages. 

The  goods,  whether  lost,  injured,  or  saved,  are  valued  at  the 
price  they  would  have  brought  in  ready  money  on  the  vessel's 
arriving  at  her  port  of  destination,  unless  the  average  is  adjust- 
ed at  the  port  of  lading,  when  they  are  valued  at  the  invoice 
price. 

In  compensating  for  the  expenses  incurred  in  repairing  dam- 
ages done  to  the  vessel,  only  §  of  the  cost  is  allowed,  as  in  gen- 
eral the  new  material  is  -J-  better  than  the  old. 

432.  To  adjust  the  general  average  of  losses  at  sea. 

Ex.  1.    The  ship  America  sailed  on  May  1G,  1857,  for  New 


GENERAL    AVERAGE.  319 

Orleans,  with  an  assorted  cargo.  In  consequence  of  a  violent 
gale  in  the  Gulf  of  Mexico,  the  captain  was  obliged  to  throw 
overboard  a  portion  of  the  cargo,  amounting  in  value  to 
$  4465.50,  and  the  necessary  repairs  of  the  vessel  cost  $  423. 
The  contributory  interests  were  as  follows :  vessel,  $  30000 ; 
gross  freight,  $  6225 ;  cargo  shipped  by  R.  S.  Davis  &  Co., 
$  3650 ;  by  Henry  Mason,  $  6500  ;  by  G.  T.  Sampson,  $  2000 ; 
by  J.  Francis  &  Son,  $550;  by  Morton  Brothers,  $5450; 
and  by  Sanborn  &  Carter,  $  8500.  Of  the  cargo  thrown  over- 
board, there  belonged  to  Henry  Mason  the  value  of  $  3000,  and 
to  Morton  Brothers  the  remainder,  $  1465.50.  The  cost  of  de- 
tention in  port  in  consequence  of  repairs  was  $  116.50.  How 
ought  the  loss  to  be  apportioned  among  the  contributory  in- 
terests ? 

OPERATION. 
CONTRIBUTORY    INTERESTS.  LOSS    FOR   GENERAL   BENEFIT. 

Vessel,  $  30000         Thrown  overboard,  $  4465.50 

Cargo,  26650         Cost  of  detention,  11 6.50 

Freight,  less  J,  4150         Repairs,  less  J,  282.00 

Entire  contrib.  int.,  $60800  Entire  loss,  $  4864.00 

$  4864.00  -r-  60800  ==  .08,  the  loss  per  cent. 
$  30000  X  .08  =  $  2400,  am't  payable  by  vessel. 

4150  X  .08  ==       332,  am't  payable  by  freight. 

3650  X  .08  =       292,  am't  payable  by  R.  S.  Davis  &  Co. 

6500  X  .08  —       520,  am't  payable  by  Henry  Mason. 

2000  X  .08  ==       160,  am't  payable  by  G.  T.  Sampson. 
550  X  .08  =c=         44,  am't  papable  by  J.  Francis  &  Son. 

5450  X  .08  ss       436,  am't  payable  by  Morton  Brothers. 

8500  X  .08  =       680,  am't  payable  by  Sanborn  &  Carter. 

Proof,  $4864,  entire  amount  payable. 

$  2400.00  —  $  398.50  =  $  2001.50,  balance  payable  by  vessel. 
3000.00—  520.00  =  2480.00,  bal.  rec'ble  by  II.  Mason. 
1465.50  —     436.00  :*=     1029.50,  bal.rec'ble  by  Morton  Bro's. 

Since  the  vessel  lost  $  116.50  -f-  $  282,  =  $  398.50,  that  amount 
is  deducted  in  finding  the  net  amount  the  vessel  must  contribute  to 
the  general  loss.  Henry  Mason  lost  $  520 ;  the  amount  payable  by 
him  is  therefore  made  so  much  less  on  that  account;  and  Morton 
Brothers  also,  having  lost  $436,  have  the  amount  of  their  contribu- 
tion lessened  by  that  sum. 


320  EQUATION   OF   PAYMENTS. 

Rule.  — Multiply  each  contributory  interest  by  the  loss  per  cent.,  and 
the  product  will  be  its  contribution  to  the  general  loss. 

Ex.  2.  The  ship  Hope,  in  her  passage  from  Liverpool  to 
New  York,  was  crippled  in  a  storm,  in  consequence  of  which 
the  captain  had  §6500  worth  of  the  cargo  thrown  overboard, 
and  the  necessary  repairs  of  the  vessel  cost  $  1050.  The 
charges  for  board  of  seamen,  pilotage,  and  dockage  amounted 
to  %  142.  The  contributory  interests  were :  vessel,  $  31500  ; 
gross  amount  of  freight,  $  4160 ;  cargo  shipped  by  Manning  & 
Brother,  $  2145  ;  by  Anderson  &  Fisk,  $  1460 ;  by  Smidt  & 
Huber,  $  960  ;  by  Greenwood,  Laporte,  &  Co.,  $  670  ;  and  by 
Allermann,  Ritter,  &  Herr,  $  1000.  In  adjusting  the  general 
average  in  New  York,  the  deduction  made  from  the  gross 
amount  of  freight  on  account  of  seamen's  wages  was  one  half. 
Required  the  several  shares  of  the  general  loss. 


EQUATION   OF   PAYMENTS. 

433.  Equation  of  Payments  is  the  process  of  finding 
the  average  or  mean  time  when  the  payments  of  several  sums, 
due  at  different  times,  may  all  be  made  at  one  time,  without 
loss  either  to  the  debtor  or  creditor. 

434.  A  strictly  accurate  method  of  determining  the  aver- 
age time  of  payment  of  several  sums  due  at  different  times 
would  require  the  finding  of  the  true  present  worth  of  the  sev- 
eral debts  at  the  given  rate  of  interest,  and  then  the  finding  in 
what  time,  at  the  same  rate,  the  sum  of  those  present  worths 
would  amount  to  the  given  debts. 

Mercantile  usage,  however,  gives  its  sanction  to  another 
method,  and  one  which  is  not  entirely  correct,  though,  when  the 
sums  are  small  and  the  terms  short,  the  variation  from  the 
exact  truth  is  practically  of  no  consequence.  But  this  method, 
being  far  the  most  convenient  of  application,  is  adopted  among 
business  men. 


EQUATION   OF   PAYMENTS.  321 

435 1  To  find  the  average  or  mean  time  of  paying  several 
debts  due  at  different  dates. 

Ex.  1.  A  owes  B  $  19,  $  5  of  which  is  to  be  paid  in  6 
months,  $  6  in  7  months,  and  $  8  in  10  months.  "What  is  the 
average  time  of  paying  the  whole  ?  Ans.  8  months. 

operation.  The    interest    of    $  5    for    6 

5  X      6  =  30  months  is  the  same  as  the  inter- 

6  X      7  =  42  est  of  $  1  for  30  months ;  and  of 
8X10  =  80  $6  for  7  months,  the  same  as  of 

S  1  for  42  months;  and  of  $ 8  for 

19  19)152(8  months.      io  months,  the  same  as  of  $  1  for 

15  2  80  months.    Hence,  the  interest  of 

all  the  sums  to  the  time  of  pay- 
ment is  the  same  as  the  interest  of  $  1  for  30mo.  -\-  42mo.  -|-  80mo.  = 
152  months.  Now,  if  $  1  require  152  months  to  gain  a  certain  sum, 
$5-f-$6-f-$8  =  $19  will  require  Jj  of  152  months;  and  152mo.  -j- 
19  =  8  months,  the  average  or  mean  time  for  the  payment  of  the  whole. 

2.  Purchased  goods  of  Kendall  &  White  at  different  times, 
and  on  various  terms  of  credit,  as  by  the  statement  annexed. 
"What  is  the  mean  time  of  payment  ? 

January    1,  a  bill  amounting  to  $  375.50  on  4  months. 


"       20, 

a                      a 

168.75  on  5  months. 

February  4, 

ti                     a 

386.25  on  4  months. 

March     11, 

a                     u 

144.60  on  5  months. 

April        7, 

a                     a 

386.90  on  3  months. 

OPERATION. 

Due  May    1, 

$3  7  5.5  0 

"    June  20, 

1  G  8.7  5  X 

50 

=     8  4  3  7.5  0 

"    June    4, 

3  8  6.2  5  X 

34 

.=  1  3  1  3  2.5  0 

"    Aug.  11, 

14  4.6  0  X 

102 

=  1  4  7  4  9.2  0 

"    July    7, 

3  8  6.9  0  X 

67 

=  2  5  9  2  2.3  0 

$  1  4  6  2.0  0  6  2  2  4  1.5  0  days. 

62241.50  -*-  1462.00  =  42$g{f  days. 

May  1  — [—  43  days  ==  June  13,  Ans. 

We  first  find  the  time  when  each  of  the  bills  will  become  due. 
Then,  since  it  will  shorten  the  operation  and  not  change  the  result, 
we  take  the  first  time  when  any  Ml  becomes  due,  instead  of  its  date,  for 
the  point  from  which  to  compute  the  average  time.  Now,  since  May 
1  is  the  period  from  which  the  average  time  is  computed,  no  time 
will  be  reckoned  on  the  first  bill,  but  the  time  for  the  payment  of  the 
second  bill  extends  50  days  beyond  May  1,  and  we  multiply  its 
amount  by  50.     Proceeding  in  the  same  manner  with  the  remaining 


322  EQUATION   OF   PAYMENTS. 

bills,  we  find  the  average  time  of  payment  to  be  43  days,  nearly,  from 
May  1,  or  on  June  13. 

Rule.  —  Multiply  each  payment  by  its  own  time  of  credit,  and  divide 
the  sum  of  the  products  by  the  sum  of  the  payments. 

Note  1.  —  When  the  date  of  the  average  time  of  payment  is  required,  as  in 
Example  2,  find  the  time  when  each  of  the  sums  becomes  due.  Multiply  each  sum 
by  the  number  of  days  intervening  between  the  date  of  its  becoming  due  and  the 
earliest  date  on  ichich  any  sum  becomes  due.  Then  proceed  as  in  the  rule,  and  the 
quotient  will  be  the  average  time  required,  in  days  forward  from  the  date  of  the 
earliest  sum  becoming  due. 

Note  2.  —  In  the  result,  it  is  customary,  if  there  be  a  fraction  of  a  day  less 
than  £,  to  reject  it;  but  if  more  than  £,  to  reckon  it  as  1  day. 

In  practice  the  work  may  be  somewhat  abridged,  without  varying  materi- 
ally the  result,  by  disregarding,  in  performing  the  multiplications,  the  cents  in 
the  several  sums,  when  they  are  less  than  50,  and  by  calling  them  $  1,  when 
more  than  50. 

When  a  payment  is  made  at  the  time  of  purchase,  it  has  no  product,  but 
it  must  be  added  with  the  other  products  in  finding  the  average  time. 

Note  3.  —  The  method  of  the  rule,  or  that  generally  adopted  by  merchants, 
as  has  been  intimated  (Art.  384),  is  not  perfectly  correct.  For  if  I  owe  a  man 
$  200,  $  100  of  which  I  am  to  pay  down,  and  the  other  $  100  in  two  years,  the 
equated  time  for  the  payment  of  both  sums  would  be  one  year.  It  is  evident 
that,  for  deferring  the  payment  of  the  first  $  100  for  one  year,  I  ought  to  pay 
the  amount  of  $100  for  that  time,  which  is  $  106;  but  for  the  $100  which  I 
pay  a  year  before  it  is  due,  I  ought  to  pay  the  present  wortli  of  $  100,  which  is 
S94.33JJ-,  and  $106  +  S  94.33JJ  =  $200.33*J;  whereas,  by  the  mercantile 
method  of  equating  payments,  I  only  pay  $  200. 

Examples. 

3.  There  is  owing  a  merchant  $  1000 ;  $  200  of  it  is  to  be 
paid  in  3  months,  $  300  in  5  months,  and  the  remainder  in  10 
months.  What  is  the  equated  time  for  the  payment  of  the 
whole  sum  ?  Ans.  7mo.  3d. 

4  I  have  bought  a  farm  for  $  6500 ;  $  2000  of  which  is  to 
be  paid  down,  $500  in  one  year,  and  the  remainder  in  2  years. 
But  if  a  note  for  the  whole  amount  had  been  preferred,  in  what 
time  would  it  have  become  due  ? 

5.  A  owes  B  $300,  of  which  $50  is  to  be  paid  in  2 
months,  $  100  in  5  months,  and  the  remainder  in  8  months. 
What  is  the  equated  time  for  the  payment  of  the  whole 
sum  ?  Ans.  6  months. 

6.  I  have  sold  H.  W.  Hathaway  several  bills  of  goods,  at 
different  times,  and  on  various  terms  of  credit,  as  by  the  follow- 
ing statement.  What  is  the  average  time  for  the  payment  of 
the  whole? 


EQUATION   OF   PAYMENTS.  328 

Jan.        1,  a  bill  amounting  to  $  600,  on  4  months. 
Feb.        7,         «  "  370,  on  5  months. 

March  15,         "  "  560,  on  4  months. 

April    20,         u  "  420,  on  6  months. 

Ans.  July  12. 
7.    Purchased  goods  of  J.   D.   Martin,  at  different  times, 
and  on  various  terms  of  credit,  as  by  the  statement  annexed. 
What  is  the  equated  time  of  paying  for  the  same  ? 

March  1, 1855,  a  bill  amounting  to  $  675.25,  on  3  months. 

376.18,  on  4  months. 
"  821.75,  on  2  months. 

"  961.25,  on  8  months. 

"  144.50,  on  3  months. 

"  811.30,  on  6  months. 

"  567.70,  on  5  months. 

"  369.80,  on  4  months. 

Ans.  March  15,  1856. 

4S6.  To  find  the  time  of  paying  the  balance  of  a  debt, 
when  partial  payments  have  been  made  before  the  debt  is 
due. 

Ex.  1.  I  have  bought  of  Leonard  Johnson  goods  to  the 
amount  of  $  1728,  on  6  months'  credit.  At  the  end  of  one 
month  I  pay  him  $  300,  and  at  the  end  of  5  months,  $  800. 
How  long,  in  equity,  after  the  expiration  of  6  months,  should 
the  balance  remain  unpaid  ?  Ans.  3mo.  20d. 

operation.  The  interest  on  the  $300 

300x5  =  1500  for  5  months  is  equal  to  the 

q  0  0  v   1  oaa  interest  of  $  1  for  1500  months, 

and  the  interest  of  $  800  for.l 


July    4,     " 

a 

Sept.  25,     " 

a 

Oct.      1,     " 

Ci 

Jan.      1, 1856, 

a 

Feb.   10,     " 

u 

Mar.  12,     " 

U 

April  15,     " 

a 

1100       •  2300  month  is  equal  to  that  of  $  1 

$1  7  2  8  —  $1  1  0  0  =  $6  2  8;    ^  80°  m0If^  ^J*™  the 
T  interest  on  both  partial  pay- 

2300  —  628  =  3mo.  20d.       ments,  at  the  expiration  of  the 

6  months,  is  equal  to  the  in- 
terest of  $  1  for  1500  -f-  800  =  2300  months.  To  equal  this  credit 
of  interest,  the  balance  of  the  debt,  which  we  find  to  be  $  628,  should 
remain  unpaid,  after  the  6  months,  -^  of  2300  months,  or  3  months 
and  20  days. 

Rule.  —  Multiply  each  payment  by  the  time  it  teas  made  before  it  be- 
comes due,  and  divide  the  sum  of  the  products  by  the  balance  remaining 
unpaid ;  and  the  Quotient  ivill  be  the  required  time. 


324  EQUATION   OF   PAYMENTS. 

Examples. 

2.  A  merchant  has  $  144  due  him,  to  be  paid  in  7  months, 
but  the  debtor  agrees  to  pay  one  half  ready  money,  and  two 
thirds  of  the  remainder  in  4  months.  What  time  should  be 
allowed  for  paying  the  balance  ?  Ans.  2y.  10mo.' 

3.  March  23,  1856,  I  sold  John  Morse  goods  to  the  amount 
of  $  8000  on  a  credit  of  8  months.  April  5,  he  paid  me  $  1200  ; 
July  4,  $1500;  September  25,  $1800;  October  1,  $1000; 
November  20,  $  500.  "When,  in  equity,  should  I  receive  the 
balance  ? 

4.  There  is  due  to  a  merchant  $  800,  one  sixth  of  which  is 
to  be  paid  in  2  months,  one  third  in  3  months,  and  the  remain- 
der in  6  months ;  but  the  debtor  agrees  to  pay  one  half  down. 
How  long  may  the  debtor  retain  the  other  half  so  that  neither 
party  may  sustain  loss  ?  Ans.  8§  months. 

5.  I  have  sold  Charles  Fox  goods  to  the  amount  of  $  3051, 
on  a  credit  of  6  months,  from  September  25,  1856.  Octo- 
ber 4,  he  paid  %  176  ;  November  12,  $375;  December  5, 
$800;  January  1,  1857,  $200.  When,  in  equity,  ought  I  to 
receive  the  balance  ?  .  Ans.  October  8,  1857. 

AVERAGING  OF  ACCOUNTS. 

,  437*  An  Account  Current  is  a  statement  of  the  mercan- 
tile transactions  of  one  person  with  another,  when  immediate 
payments  are  not  made. 

An  account  is  marked  Dr.  on  the  left,  to  indicate  that  the 
person  with  whom  the  account  is  kept  is  debtor  for  the  items  on 
that  side ;  and  is  marked  Cr.  on  the  right,  to  indicate  that  he  is 
creditor  for  the  items  on  that  side. 

Accounts  current  are  generally  made  up  or  settled  at  the  end 
of  every  six  months  or  year. 

438.  To  find  the  equated  time  when  the  balance  of  an 
account  current  will  be  due. 

Ex.  1.  In  the  following  account  when  did  the  balance  be- 
come due,  the  merchandise  articles  being  on  6  months'  credit  ? 

Ans.  December  22,  1856. 


EQUATION    OP    PAYMENTS. 


325 


Dr. 

Messrs.  James  Button  «jp  Co.  in  account  tvith  David  Hale.      Cr. 

1856. 

1 

1856. 

Jan.  4. 

To  merchandise, 

$  96.51 

Jan.30. 

By  cash, 

$240.00 

"    18. 

U                    M 

57.67 

Apr.  3. 

«        u 

48.88 

Feb.  4. 

"  cash  paid  draft, 
"  merchandise, 

80.00 

May22. 

U          U 

50.00 

u      a 

38.96 

July  7. 

"    Note,1*  June  22,  6mo. 

410.01 

"     9. 

"  cash  paid  draft, 

50.26 

/ 

Mar.  3. 

"  merchandise, 

154.46 

/ 

"  24. 

u             « 

42.30 

/ 

Apr.  9. 

u             u 

23.60 

/ 

May  15. 

«               u 

28.46 

/ 

"  21. 

u                a 

177.19 

/ 

$748.89 

$748.89 

Errors  excepted. 


Debits. 
Due  Feb.  4,    80 


Settled  as  above,  Boston,  July  7,  1856. 

David  Hale, 
By  John  Davis. 

FIRST    OPERATION. 

Credits. 
Due  Jan.  30,  240 
"    April  3,    49  X    65  =  3136 
"    May  22,    50X113  =  5650 

$  339  8786  days. 

8786  -r  339  =  25j|i  days. 

Credits  due  26  days  from  January  30, 
or  on  February  25. 

Difference  between  February  25  and 
August  7  =  164  days. 

$  749  —  $  339  =  $  410,  balance. 
164  X  339  =  55596; 
55596  -^  410  =  135123  days. 

136  days  forward  from  August  7, 1856  =  December  22, 1856. 

On  equating  each  side  of  the  account  (Art.  435),  we  find  the  debits 
became  due  186  days  from  February  4,  or  on  August  7 ;  and  the 
credits  became  due  26  days  from  January  30,  or  on  February  25. 

If  the  account  had  been  settled  on  February  25,  it  is  evident  the 
debits  would  have  been  paid  164  days,  or  the  time  from  February  25 
to  August  7,  before  having  become  due.  This  would  have  been  a 
loss  of  interest  to  the  debit  side  of  the  account,  and  a  corresponding 
gain  to  the  credit  side.  Now,  as  the  settlement  should  be  one  of 
equity,  we  find  how  long  it  will  take  the  balance,  $410,  to  gain  the 
same  interest  that  %  339  would  gain  in  the  164  days.  If  it  take  $  339 
to  gain  a  certain  interest  in  164  days,  it  would  take  %  1  to  gain  the 
same  interest  339  times  164  days  =  55596  days;  and  it  would  take 
$410  to  gain  the  same  amount  of  interest  T^  of  55596  days  =  136 


u 

Feb.  9,    50  X      5  =      250 

II 

July  4,    97  X  151  =  14647 

Ij 

"   18,    58  X  165  =    9570 

» 

Aug.  4, 154  X  212  =  32648 

a 

"    24,   42  X  233  =    9786 

a 

"      9,   24  X  248  =    5952 

M 

Nov.15,  28  X  285  =    7980 

14 

"    21,177  X  291  =  51509 

$  749                139438  days. 

139438  -7-  749  =  18632A  days. 

el 

ts  due  186  days  from  February  4, 

or  on  August  7. 

*  Included  only  to  illustrate  the  manner  of  settling  an  account. 
N        2§ 


Debits. 

Due  Feb.  4, 

80  X   5  = 

400 

M 

Feb.  9, 

50  X  10  = 

500 

a 

July  4, 

97  X  156  = 

15132 

a 

"  18, 

58  X  169  = 

9802 

u 

Aug.  4, 

39  X  187  = 

7293 

(( 

Sept.  3, 

154  X  217  = 

33418 

M 

"  24, 

42  X  238  = 

9996 

M 

Oct.  9, 

24  X  253  = 

6072 

U 

Nov.  15, 

28  X  290  = 

8120 

a 

"  21, 

177  X  296  = 

52392 

326  EQUATION   OF   PAYMENTS. 

days  nearly.    Hence,  the  balance  became  due  136  days  forward  from 
August  7,  1856,  or  on  December  22,  1856. 

The  time  was  counted  forward  from  the  average  date  of  the  larger 
amount,  since  it  became  due  last ;  but  had  that  amount  become  due 
first,  the  time  would  have  been  counted  backward  from  its  average 
time. 

SECOND    OPERATION. 

Credits. 
Due  Jan.  30,  240 
"    April  3,    49  X    65  =  3136 
"    May  22,  J50  X  113  =  5650 

$  339  8786  days. 

$  749  —  $  339  =  $  410,  balance  of  the 

items. 
143125  —  8786  =  134339,  balance  of 
products. 

134339  -r  410  =  327-2  5  0  days. 
$  749  143125  days. 

328  days  forward  from  January  30, 1856  =  December  22,  1856. 

In  the  second  operation,  we  take  the  earliest  date  on  which  any 
sum  becomes  due  in  the  account,  for  the  starting-point  from  which 
to  reckon  the  days,  by  which  to  find  the  several  products  belonging 
either  to  the  debit  or  credit  side  (Art.  435,  Note  1).  The  sum  of  the 
debit  products,  143125,  denotes  the  number  of  days  required  for  $  1 
to  gain  as  much  interest  as  all  the  items  of  debit  would  gain  in  the 
times  of  their  becoming  due,  and  the  sum  of  the  credit  products,  8786, 
denotes  the  number  of  days  required  for  $  1  to  gain  as  much  interest 
as  all  the  items  of  debit  would  gain  in  the  times  of  their  becoming  due. 
The  difference  between  the  sums  of  debit  and  credit  products  is 
134339,  and  the  difference  between  the  debit  and  credit  items  is 
$410.  Then,  if  it  requires  134339  days  for  $  1  to  gain  a  certain 
interest,  it  will  require  $410  to  gain  the  same  amount  ^i^  of  134339 
days  =328  days  nearly.  328  days  forward  from  January  30,  1856, 
=  December  22, 1856,  the  time  of  the  balance  of  the  account  becom- 
ing due,  as  before. 

Rule  1.  —  Find  the  average  time  of  each  side  becoming  due. 

Multiply  the  amount  of  the  smaller  side  by  the  number  of  days  between 
the  two  average  dates,  and  divide  the  product  by  the  balance  of  the  ac- 
count. 

The  quotient  ivill  be  the  time  of  the  balance  becoming  due,  counted 
from  the  average  date  of  the  larger  side,  forward  when  the  amount 
of  that  side  becomes  due  last,  but  backward  when  it  becomes  due 
first.     Or, 

Rule  2.  — Multiply  each  sum  of  debit  and  credit  by  the  number  of 
days  intervening  between  the  date  of  its  becoming  due,  and  the  earliest 
date  on  which  any  sum  in  the  account  becomes  due.  . 

Then,  the  difference  between  the  sums  of  debit  and  credit  products, 
divided  by  the  difference  between  the  debit  and  credit  item,  will  give  the 


EQUATION   OB'   PAYMENTS. 


327 


time,  to  be  counted  from  the  earliest  date  of  any  sum  in  the  account  be- 
coming due,  forward  when  the  larger  sum  of  products  is  on  the  larger 
side  of  the  account,  but  backward  when  it  is  on  the  smaller  side. 

Note.  —  The  cash  value  of  a  balance  of  an  account  drawing  interest,  or 
whose  items  are  on  different  terms  of  credit,  depends  upon  the  time  of  settle- 
ment, and  is  therefore  either  larger  or  smaller  than  the  difference  between  the 
debit  and  credit  items. 

The  average  time  of  a  balance  becoming  due  being  known,  its  cash  value 
may  be  found,  when  the  balance  is  due  before  the  time  of  settlement,  by  adding 
to  the  balance  interest  up  to  the  time  of  settlement;  and  when  due  after  that 
time,  by  deducting  from  the  balance  interest  for  the  time  intervening  between"  the 
time  of  settlement  and  the  time  of  the  balance  becoming  due.  The  deduction  of 
interest,  in  the  latter  case,  is  the  mercantile  method,  instead  of  that  of  finding 
the  true  present  worth  by  deducting  the  true  discount. 


Examples. 

2.  Required  the  time  when  the  balance  of  the  following  ac- 
counts becomes  subject  to  interest,  allowing  that  each  item  was 
due  from  its  date. 
Dr.  D.  Wadsworth  in  account  with  S.  Adams.  Cr. 


1855. 

1855. 

July     4, 

To  balance, 

$  375.90 

Aug.  10, 

By  cash, 

$316.00 

Aug.  20, 

"  merchandise, 

815.58 

Sept.    1, 

a      u 

675.00 

"     29, 

u             « 

178.25 

"     25, 

"  merchandise, 

512.25 

Sept.  25, 

u             u 

387.20 

Nov.  20, 

"  cash, 

161.75 

Dec.     5, 

u              u 

418.70 

Dec.     1, 

U       a 

100.00 

Ans.  August  5,  1855. 

3.  When  did  the  balance  of  the  following  account  become 
due,  the  merchandise  items  being  on  6  months  ? 

Dr.  John  Greene  in  account  with  F.  Johnson.  Cr. 


1856. 

1856. 

March  1, 

To  merchandise, 

$  720.75 

April    1, 

By  cash, 

$  700.00 

"     20, 

M 

u 

815.30 

May  30, 

"  merchandise, 

569.89 

April  11, 

<l 

u 

587.80 

July  20, 

"  cash, 

500.00 

"    30, 

u 

u 

300.00 

Sept.  25, 

U          li 

100.00 

June  15, 

a 

u 

625.25 

"     30, 

"  merchandise, 

750.20 

July  18, 

a 

(C 

560.00 

Oct.   30, 

14                U 

329.96 

Aug.  30, 

n 

u 

684.90 

Nov.  20, 

U                  M 

500.00 

Sept.  25, 

u 

u 

365.30 

4.  Allowing  that  each  item  of  the  following  account 
draws  interest  from  its  date,  at  what  time  would  the  balance 
become  due,  and  how  much  ready  money  should  in  equity 
discharge  the  same,  September  21,  1856,  interest  being  at  6 
per  cent.  ? 


328 


EQUATION   OF   PAYMENTS. 


Dr. 


1.  Bradley  in  account  ivith  T.  B.  Fuller. 


Cr. 


1856. 

1856. 

March  1, 

To  merchandise, 

$  36.25 

April    1, 

By  cash, 
u       u 

$  48.25 

April    7, 

M                  (t 

18.15 

May  20, 

90.10 

June  15, 

U              a 

48.26 

June  17, 

U      u 

12.50 

July  21, 

u              u 

91.20 

July     4, 

u       c( 

20.00 

Aug.    1, 

ti             u 

30.00 

"     10, 

«       a 

25.00 

Ans.  November  21, 1856  ;  cash  value  of  balance,  $  27.73. 

5.  Required  the  time  when  the  balance  of  the  following  ac- 
count became  subject  to  interest,  allowing  the  merchandise  to 
have  been  on  8  months ;  and  the  cash  value  of  the  balance  on 
November  28, 1857,  provided  it  drew  6  per  cent,  interest  from 
the  time  of  its  becoming  due. 
Dr.  N.  Chandler,  2d,  in  account  ivith  T.  E.  Lanman.  Cr. 


1856. 

1857. 

May     1, 

To  merchandise, 

$  300.00 

Jan.      1, 

By  cash, 

"  merchandise, 

$500.00 

July    7, 

u           u 

759.96 

Feb.   18, 

481.75 

Sept.  11, 

a           n 

417.20 

Mar.  19, 

"  cash, 

750.25 

Nov.  25, 

U                   U 

287.70 

April    1, 

"  draft, 

210.00 

Dec.  20, 

tl                   a 

571.10 

May  25, 

"  cash, 

100.00 

Ans.  July  28, 1857  ;  cash  value  of  balance,  $  299.88. 

6.  The  following  account  was  settled  May  24,  1857 ;  how 
long  previous  to  date  was  the  balance  by  average  due,  and  what 
was  the  cash  value  of  the  balance  at  the  time  of  settlement  ? 


Dr. 


David  Taggart  in  account  with  George  Perry. 


Cr. 


1856. 

1856. 

Jan.      1, 

To  m'dise  on  5mo., 

$560.00 

March  7, 

By  m'dise  on  6mo., 

$350.00 

Feb.   11, 

"        "      on  6mo., 

846.00 

April  17, 

a           u           a         u 

820.00 

Mar.  20, 

"        "      on  6mov 

728.00 

June  20, 

"   cash, 

100.00 

July  30, 

"  cash, 

400.00 

Aug.  15, 

u        u 

800.00 

Sept.  12, 

"  m'dise  on  3mo., 

560.00 

Sept.  18, 

"   m'dise  on  6mo., 

630.00 

Dec.  18, 

"  cash, 

600.00 

Oct.   28, 

"    cash, 

400.00 

1857. 

Nov.     1, 

"  m'dise  on  4mo., 

750.00 

May  10, 

If          K 

500.00 

Ans.  May  7,  1856;  cash  value  of  balance,  $365.61, 


439.  To  find  the  true  balance  of  an  account  current  whose 
items  draw  interest. 

Ex.  1.  Required  the  true  balance  of  the  following  account, 
on  November  1,  1857,  the  time  of  settlement,  allowing  that 
each  item  drew  interest  from  its  date,  at  the  rate  of  6  per  cent. 

Ans.  $  430.04. 


EQUATION   OF   PAYMENTS. 


329 


Dr. 


Wesley  Scott  in  account  with  SetJi  T^son. 


Cr. 


1857. 

f    1857. 

Jan.    17, 

To  merchandise, 

$  144.05 

Feb.     1, 

By  cash, 

$200.00 

Mar.  20, 

(1                u 

374.95 

March  9, 

U        u 

300.00 

May     7, 

u            U 

500.00 

April    1, 

u        u 

600.00 

July    4, 

"  cash, 

100.00 

"       5, 

"  merchandise, 

180.00 

"       7, 

"  merchandise, 

600.00 

June  12, 

a             u 

700.00 

Sept.  25, 

it               u 

250.00 

Aug.  19, 

"  cash, 

400.00 

Nov.     1, 

"  *bal.  new  acc't, 

430.04 

Nov.    1, 

"  ^balance  of  int., 

19.04 

$  2399.04 

$  2399.04 

Errors  excepted.        Philadelphia,  November  1, 1857. 


Seth  Wilson. 


OPERATION. 


Debits. 
Due  Jan.      17,  144  X  288  =  44472 
March  20,  375  X  226  ==  84750 
May       7,  500  X  178  =  89000 
July      4,  100  X  120  =  12000 
"  7,  600  X  117  =  70200 

Sept.    25,  250  X     37  =    9250 


Credits. 


Due  Feb.  1,  200  x  273  =  54600 
"  March  9,  300  X  237  =  71100 
April  1,  600  X  214  =  128400 
"  5,  180  X  210  =  37800 
"  June  12,  700  X  142  =  99400 
"  Aug.  19, 400  X  74=  29600 


a 


u 


$1969 


6  )  306.672 


2380 


6  )  420.900 
$  70.150 


$51,112 

$  2380  —  $  1969  =  $  411,  balance  of  items. 
$  70.15  —  $  51.11  =  $  19.04,  balance  of  interest. 
$411  +  $  19.04  =  $  430.04,  true  balance. 

The  time  in  days  intervening  between  the  date  of  each  item  and 
the  time  of  settlement  is,  evidently,  the  number  of  days  each  item  is 
on  interest.  Then,  if  each  item  be  multiplied  by  the  number  denot- 
ing its  number  of  days,  and  divided  by  6000,  the  result  will  be  its 
interest  (Art.  356,  Note  1),  and  the  sum  of  the  interest  of  the  several 
items  of  debit  will  be  the  aggregate  interest  of  the  debit  side,  and  the 
same  principle  will  hold  in  finding  the  aggregate  interest  of  the  credit 
side.  But  the  same  result  may  be  obtained  in  a  shorter  way  by  di- 
viding the  sums  of  the  debit  and  credit  products  by  6000,  which  is 
done  most  readily  by  pointing  off  three  decimal  places  at  the  right  of 
each  sum  (Art.  78),  and  dividing  by  6,  as  in  the  operation.  Then 
the  balance  of  items  and  the  balance  of  interest  we  add  together,  and 
thus  obtain  the  true  balance. 

2.  Required  the  true  balance  of  the  following  account,  on 
October  1,  1857,  the  time  of  settlement,  allowing  the  rate  of 
interest  to  be  6  per  cent.  Ans.  $  290.77. 


*  Introduced  only  to  illustrate  the  manner  of  balancing  the  account. 

28* 


330 


EQUATION   OF   PAYMENTS. 


Dr. 


Grant  $f  Barker  in  account  and  interest  with  J.  Ritchie.        Cr. 


1857. 

1857. 

Mar.   1, 

To  balance  of  old  acc't. 

$100.15 

Mar.  15, 

By  m'dise  on  6mo., 

$  160.00 

"     15, 

"  cash  paid  draft, 

40.00 

April  1, 

"  cash, 

100.00 

May  20, 

"  merch'dise  on  6mo.; 

180.85 

June   1, 

"  draft, 

120.00 

Oct.    1, 

"  ^balance, 

290.77 

Aug.  10, 

"  m'dise  on  2mo., 

80.00 

j^ 

Sept.  1, 

"  cash  paid  draft, 

50.00 

// 

Oct.    1, 

"  ^balance  of  int., 

.77 

$  610.77 

$  610.77 

Debits. 
Due  March    1,  100  x  214  = 
"         "      15,     40  X  199  = 
"    Sept.    20,  180  X    11  = 


OPERATION. 

Debit  products. 

21400 

7960 

1980 


Credit  products. 


$320 

31340 

Credits. 

ue 

April 

1,  100  X  183  = 

18300 

u 

June 

1,  120  X  122  = 

14640 

M 

Sept. 

1,     50  X     30  = 

1500 

U 

v 

15,  160  X     14  = 

2240 

u 

Oct. 

10,     80  X       9  = 

720 

$610 
320 

$  290  balance  of  items. 


82060 


36680 
32060 

6  )  4.620 

$  0.77  bal.  of  interest. 


$290  +  $0.77  =  $290.77,  true  balance. 

We  find  the  several  products  by  multiplying,  as  in  working  the 
preceding  example.  The  products  belonging  to  items  becoming  due 
previous  to  settlement  we  arrange  as  belonging  to  their  respective  sides. 
One  item,  however,  becomes  due  9  days  after  the  time  of  settlement, 
thus  requiring  its  side  to  be  diminished  by  the  interest  of  that  item 
for  the  9  days,  or  that  the  opposite  side  should  be  increased  by  that 
interest.  We,  therefore,  write  the  product  of  this  item  with  the 
debit  products.  We  may  now  find  the  interest  of  each  side  sepa- 
rately, and  then  subtract  that  of  the  one  side  from  that  of  the  other 
for  the  balance  of  interest;  but  we  obtain  the  desired  result  by  a 
shorter  method  by  finding  the  interest  corresponding  to  the  differ- 
ence of  the  sums  of  the  debit  and  credit  products.  Hence  the 
general 

Rule.  —  Multiply  each  item  of  debit  and  credit  by  the  number  of 
days  intervening  between  its  becoming  due  and  the  time  of  settlement. 

Place  the  product  of  each  item  becoming  due  before  the  time  of  set- 
tlement on  its  own  side  of  the  account,  and  place  the  product  of  each  item 


*■  Introduced  only  to  illustrate  the  manner  of  settlement. 


EQUATION   OF   PAYMENTS.  331 

becoming  due  after  the  time  of  settlement  on  the  side  opposite  to  its 
own. 

The  difference  of  the  sums  of  the  debit  and  credit  products,  with  three 
places  pointed  off  on  the  right  for  decimals,  divided  by  6,  will  give  the 
balance  of  interest 

Place  the  balance  of  interest  on  its  own  side  of  the  account,  and  the 
difference  then  between  the  two  sides  will  be  the  true  balance. 

Note  1.  —  The  interest  as  found  by  the  rule  is  that  at  6  per  cent.  From 
which  the  interest  at  any  other  rate  may  be  found  by  aliquot  parts. 

Note  2.  —  Besides  the  above  method  of  settling  accounts  drawing  interest, 
accountants  often  make  use  of  tables  constructed  to  aid  in,  equating  and 
balancing  accounts. 

Examples. 

3.  Alfred  Hicks  is  in  account  and  interest  with  Keen  & 
Lee,  as  follows :  Debtor,  January  1, 1857,  to  merchandise,  on  6 
months,  $  156.10  ;  February  3,  to  cash  paid  draft,  $  100 ; 
March  20,  to  merchandise,  on  4  months,  $316.90;  March  30, 
to  merchandise,  on  4  months,  $162.00;  May  15,  to  cash  paid 
draft,  $  100 ;  August  20,  to  merchandise,  on  6  months,  $  213.00. 
Creditor,  February  1,  by  cash,  $  120.00 ;  March  20,  by  mer- 
chandise, on  4  months,  $  420.16 ;  May  1,  by  merchandise,  on  6 
months,  $  300 ;  July  1,  by  merchandise,  on  4  months,  $  50 ; 
September  10,  by  merchandise,  on  4  months,  $  99.84.  Re- 
quired the  true  balance,  if  settled  on  December  1,  1857, 
interest  being  at  6  per  cent.  Ans.  $  61.29. 

4.  Kequired  the  true  balance  due  on  the  following  account, 
on  March  25, 1857,  each  item  drawing  7  per  cent,  interest  from 
its  date.  Benjamin  Lyman  in  account  and  interest  with  John 
Russell:  Debtor,  July  4,  1856,  to  merchandise,  $200;  Sep- 
tember 8,  to  merchandise,  $  300 ;  September  25,  to  merchan- 
dise, $  250 ;  October  1,  to  merchandise,  $  600 ;  November  20, 
to  merchandise,  $400;  December  12,  to  merchandise,  $500; 
January  15,  1857,  to  merchandise,  $100;  March  11,  to  mer- 
chandise, $120.  Creditor,  July  20,  1856,  by  cash,  $300; 
August  15,  by  cash,  $350;  September  1,  by  cash,  $400; 
November  1,  by  cash,  $  320 ;  December  6,  by  merchandise, 
$  600  ;  December  20,  by  cash,  $  100 ;  February  1,  1857,  by 
cash,  $  200  ;  February  28,  by  merchandise,  $  150. 

Ans.  $  50.64. 


332  ACCOUNTS  OF  STORAGE. 


ACCOUNTS   OF   STORAGE. 

440«  Accounts  of  storage  of  property  contain  an  entry 
of  the  number  of  articles  received  and  delivered,  with  the  date 
of  each  transaction. 

Storage  is  usually  reckoned  by  the  month  of  30  days,  at  a 
certain  price  per  barrel,  bale,  box,  &c. 

The  number  of  articles  on  which  storage  is  chargeable  for 
one  month,  or  any  other  time  agreed  upon,  is  usually  deter- 
mined by  an  average. 

44 1.  To  find  the  average  of  storage  for  a  month,  or  any 
other  time. 

Ex.  1.  What  will  be  the  cost  for  the  storage  of  flour  at  6 
cents  per  barrel,  which  was  received  and  delivered  as  follows : 
Received  May  1,  1857,  1000  barrels ;  May  26,  2000  barrels. 
Delivered  May  16,  500  barrels;  June  1,  1000  barrels;  June 
12,  1100 ;  July  12,  400.  Ans.  $  104. 

operation.  The    storage    of 

1857.  bbl.  d.  prod.         lOOObbl.     for    15d. 

May    1,  Rec.    1000x15  =  15000 


16,  Deliv.     5  00  —  2500bbl.  for  5d. 

loOObbl.  for  lid. 


Bal.        500x10=     5000 


500bbl.  for  lOd. 


400bbl.  for  20d., 


"     26,  Rec.    2000  is  the  same  as  the 

Bal.     2  5  00  X      5=     7  5  0  0  ZlT  fruXl' 

June    1,  Deliv.  1000  for  a  month  of  30 

Bal.     1500X11  =  16500  days     And  the  stor- 

"    12,  Deliv.1100  ageofl733ibbl.  at 

'               6  cents  each  equals 

Bal.        400x20=      8000  $104,  the    answer 

~" reouircd 

July    2,  Deliv.     40  0           310)520010  ln  practIce)  it  is 

Chargeable  for  1  month,  1  7  3  3  A    customary  when  the 

J    number  of   articles 
l/33^bbl.  X  .06=  $104,  cost  of  storage,    upon    which     stor- 
age is  chargeable,  as 
found,  contains  a  fraction  less  than  a  half,  to  reject  the  fraction ;  but  if 
it  is  more  titan  a  half,  to  regard  it  as  an  entire  article. 

Rule.  —  Multiply  the  number  of  barrels,  or  other  articles,  by  the 
number  of  days  they  are  in  store,  and  divide  the  sum  of  the  products 
by  30,  or  the  number  of  days  in  any  term  agreed  upon.  TJie  quotient 
will  give  the^  number  of  barrels,  or  other  articles,  on  which  storage  is 
chargeable  for  that  term. 


MISCELLANEOUS   EXAMPLES.  833 

Examples. 

2.  What  is  the  cost  of  storage  of  tea  at  3  cents  a  chest  per 
month,  received  and  delivered  as  follows:  Received,  May  16, 
1857,  4560  chests.  Delivered,  May  30,  564  chests;  June  1, 
904  chests  ;  July  9,  1000  chests  ;  August  3,  1500  chests  ;  and 
August  16,  the  balance.?  1_  t|  *j »  tyl 

3.  Received,  and  delivered,  on  account  of  Richard  Gordon, 
sundry  bales  of  cotton,  as  follows :  Received,  January  1,  1857, 
2310  bales;  January  16,  120  bales;  February  1,  300  bales. 
Delivered,  February  12,  1000  bales;  March  1,  600  bales; 
April  3,  400  bales ;  April  10,  312  bales.  Required  the  num- 
ber of  bales  remaining  in  store  on  May  1,  and  the  cost  of 
storage  up  to  that  date,  at  the  rate  of  5  cents  a  bale  per 
month.  Ans.  In  store,  418  bales  ;  cost,  $  306.90. 


MISCELLANEOUS  EXAMPLES. 

1.  The  stocks  of  three  partners,  A,  B,  and  C,  are  $3500, 
$  2200,  and  $  2500,  and  their  gains  $  1120,  $  880,  and  $  1200, 
respectively,  and  B's  stock  continued  in  trade  2  months  longer 
than  A's.     Required  how  long  the  money  of  each  was  in  trade. 

Ans.  A's  money,  8mo. ;  B's,  lOmo. ;  C's,  12mo. 

2.  A  merchant  failed  for  $  15000.  On  settling  up,  his  net 
assets,  equably  distributed,  gave  only  $  540  to  a  creditor  whose 
demand  was  $  660  more  than  that  sum.  How  much  did  the 
bankrupt  pay  on  a  dollar,  and  how  much  did  he  owe  more  than 
he  could  pay  ? 

Ans.  45  cents  on  a  dollar ;  owed  $  8250  more  than  he  could 
pay. 

3.  A  and  B  pay  a  poll  tax  each  of  $  1.50,  and  a  property 
tax  at  the  rate  of  7  mills  on  a  dollar.  A's  entire  tax  is  $  64.50, 
which  is  just  $  14  more  than  B's  entire  tax.  What  is  the  tax- 
able property  of  each  ? 

Ans.  A's  property,  $  9000 ;  B's  property,  $  7000. 

4.  The  expenses  of  a  district  school  are,  for  fuel,  $  20,  for 


334  MISCELLANEOUS   EXAMPLES. 

repairs  of  the  school-house  $  30,  and  for  teacher's  salary  $  150, 
and  the  public  money  amounts  to  $  50.  If  the  rate  bills  re- 
quire to  be  made  out  at  the  rate  of  3  cents  a  day  for  each 
pupil's  attendance,  what  was  the  aggregate  attendance  ? 

Ans.  5000  days. 

5.  J.  Kimball  had  goods  to  the  value  of  $  7000  on  board  a 
vessel,  which,  from  stress  of  weather,  required  cargo  to  the 
amount  of  $  4000  to  be  thrown  .overboard,  and  of  which  the 
value  of  $3000  belonged  to  Kimball.  If,  in  adjusting  the 
general  average,  the  several  contributory  interests  pay  5  per 
cent.,  to  what  sum  will  Kimball's  loss  be  reduced  ? 

Ans.  $  350. 

6.  A  owes  B  $  150,  $  50  to  be  paid  in  4  months,  and  $  100 
in  8  months.  B  owes  A  $  250  to  be  paid  in  10  months.  It  is 
agreed  between  them  that  A  shall  make  present  payment  of  his 
whole  debt,  and  that  B  shall  pay  his  so  much  sooner  as  to  bal- 
ance the  favor.  Required  the  time  at  which  B  must  pay  the 
$  250.  Ans.  6  months. 

7.  A  debt  is  to  be  paid  £  down,  £  in  6  months,  £  in  8  months, 
and  the  balance  in  12  months.  If  the  payments  were  all  con- 
verted into  one,  on  what  credit  should  it  be  ?    Ans.  5mo.  25d. 

8.  A  merchant  proposes  to  admit  a  young  man  into  business 
with  him,  on  condition  that,  if  he  put  into  the  stock  $  2000,  his 
pay  shall  be  $  800  a  year,  or  if  he  put  in  $  4000,  he  shall  have 
$  1100  a  year;  in  this  offer  what  was  allowed  for  his  services 
only  ?  Ans.  $  500. 

9.  A  gentleman  is  owing  three  notes  to  George  Shannon,  one 
of  $  100  due  in  4  months,  another  of  $  100  due  in  8  months, 
and  a  third  of  $  200  due  in  12  months.  Should  the  three  notes 
be  converted  into  two  for  the  same  amount,  the  one  to  run  just 
twice  as  long  as  the  other,  when  ought  they  to  be  made  pay- 
able ?  Ans.  The  one  in  6mo.,  and  the  other  in  12mo. 

10.  An  account  settled  January  1,  1858,  showed  a  balance 
of  debts  to  the  amount  of  $360,  and  a  balance  of  interest  in 
favor  of  the  credit  side  to  the  amount  of  $  3.78  ;  how  long  in 
equity  ought  the  balance  of  debts  to  remain  unpaid  after  the 
day  of  settlement,  exactly  to  offset  this  balance  of  interest, 
allowing  the  rate  of  interest  to  be  6  per  cent.  ?    Ans.  63  days. 


INSUEANCE.  335 


INSURANCE. 

442.  Insurance  is  a  contract  of  indemnity,  by  which  one 
party  engages,  for  a  stipulated  sum,  to  insure  another  against  a 
risk  or  loss  to  which  he  is  exposed. 

The  insurer  or  underwriter  is  the  party  taking  the  risk. 

The  premium  is  the  sum  paid  for  insurance,  and  is  generally 
reckoned  at  a  certain  per  cent,  of  the  value  of  the  property  in- 
sured. 

The  policy  is  the  written  obligation  or  contract. 

Note.  —  Insurance  is  generally  made  by  an  incorporated  joint-stock  com. 
pany,  and  sometimes  by  individuals.  When  each  person  insured  becomes  a 
member  and  proprietor  in  the  profit  or  loss  of  the  concern,  it  is  called  a  mutual 
insurance  company.  Many  companies,  as  a  security  against  fraud,  do  not  in- 
sure property  for  its  full  value. 

FIRE  AND  MARINE  INSURANCE. 

443 1  Fire  Insurance  is  that  which  indemnifies  damage  and 
loss  caused  by  fire  or  lightning. 

444 •  Marine  Insurance  is  that  which  indemnifies  damage 
and  loss  caused  by  the  perils  peculiar  to  navigation. 

445»  To  compute  the  premium  of  insurance  on  any  given 
amount  at  a  specified  rate. 

Ex.  1.  What  would  be  the  premium  for  insuring  a  house, 
valued  at  $  5728,  at  If  per  cent.  ?  Ans.  $  100.24. 

OPERATION. 

$5728  X  .01 J  =  $100.24. 

Rule.  —  Find  the  percentage  of  the  given  sum  at  the  rate  of  insur- 
ance, and  the  result  is  the  premium. 

Note.  —  When  the  amount  insured  and  the  premium  are  known,  the  rate  of 
insurance  may  be  found  by  Art.  350. 

Examples. 

2.  What  is  the  premium  for  insuring  a  house  valued  at 
*$  896,  at  12  per  cent.  ?  Ans.  $  107.52. 

3.  How  much  would  be  required  to  be  paid  to  effect  the  in- 
surance of  a  brig  valued  at  $  17,289,  at  1£  per  cent.  ? 

Ans.  $216,111. 


336  INSURANCE. 

4.  My  ship  Keystone  State  is  valued  at  $  35000,  and  her 
cargo  at  $  75000.  I  procure  an  insurance  on  f  of  the  value 
of  the  ship,  at  3  J  per  cent.,  and  on  §  of  her  cargo,  at  2£  per 
cent.     What  is  the  amount  of  premium  ?        Ans.  $  1932.50. 

5.  My  library,  worth  $  3675,  I  got  insured  by  paying  4J  per 
cent. ;  and  the  policy  cost  me  $  1.  The  library  having  been 
destroyed  by  fire,  what  was  my  actual  loss  ?   Ans.  $  180.15  J. 

6.  The  premium  for  insuring  $  9870  was  $  690.90.  ,  What 
was  the  rate  per  cent.  ? 

7.  White  &  Bigelow  effect  an  insurance  on  their  store  and 
goods,  worth  $47600,  for  5  years.  The  first  year  they  are  to 
pay  4£  per  cent. ;  the  second  year,  3  J  per  cent. ;  the  third  year, 
4§  per  cent. ;  the  fourth  year,  5  per  cent. ;  and  the  fifth  year, 
5£  per  cent.  What  is  the  whole  they  are  to  pay  for  the  in- 
surance ?  Ans.  $  11,007.50. 

446i  To  find  what  sum  must  be  insured,  at  a  given  rate,  to 
cover  both  property  and  premium. 

Ex.  1.  For  what  sum  must  a  policy  be  taken  out,  to  cover 
both  property  and  premium,  the  value  of  the  property  being 
$  2475,  and  the  rate  of  insurance  10  per  cent.  ? 


OPERATION. 


|  2475  -*-  .90  =  $  2750. 

Since  the  sum  to  be  covered  by  the  insurance  includes  both  the 
property  and  the  premium,  and  as  the  premium  is  10  per  cent.,  or 
.10  of  that  sum,  the  property,  evidently,  must  be  .90  of  the  sum  for 
which  the  policy  is  to  be  taken  out. 

Rule.  —  Divide  the  value  of  the  property  to  be  covered  by  1  de- 
creased by  the  rate  of  insurance  expressed  decimally,  and  the  quotient 
will  give  the  whole  sum  to  be  covered. 

Note.  —  When  the  rate  of  insurance  and  premium  are  known,  the  amount 
insured  may  be  found  by  Art.  350. 

Examples. 

2.  A  manufacturing  company  own  a  factory  valued  at 
$26250.  For  what  sum  must  a  policy  be  taken  out  to  cover 
the  property  and  premium  of  insurance,  the  rate  being  12^  per 
cent.  ?  Ans.  $  30,000. 

3.  A  store  and  its  goods  are  worth  $  6370.  What  sum  must 
be  insured,  at  2  per  cent.,  to  cover  both  property  and  pre- 
mium ?  Ans.  $  6500. 


INSURANCE.  337 

4.  The  premium  for  insuring  a  school-house,  at  the  rate  of  1£ 
per  cent.,  was  $  50.     For  what  sum  was  it  insured  ? 

Ans.  $  4000. 

5.  If  a  policy  covering  property  and  premium  be  taken  for 
$  600,  at  10  per  cent.,  what  is  the  value  of  the  property  cov- 
ered ?  Ans.  $  540. 

6.  A  merchant  adventured  $  1000  from  Boston  to  New  Or- 
leans, at  3  per  cent. ;  thence  to  Chili,  at  5  per  cent. ;  thence  to 
Canton,  at  6  per  cent. ;  and  thence  to  Boston,  at  7  per  cent. 
For  what  sum  must  he  take  out  a  policy,  to  cover  his  adventure 
the  voyage  round  ?  Ans.  $  1241.348. 

LIFE   INSURANCE. 

447.  Insurance  on  life  is  a  contract  which  stipulates  for 
the  payment  of  a  certain  sum  of  money  on  the  death  of  one  or 
more  individuals,  in  consideration  of  an  immediate  payment,  or 
an  annual  premium,  being  made  by  the  insured. 

A  temporary  insurance  on  life  is  a  contract  to  pay  a  certain 
sum  in  case  a  given  individual  dies  within  a  given  number  of 
years. 

448.  The  amount  of  premium  required  of  the  insured  as  a 
security  for  the  payment  of  a  certain  sum  at  his  death  by  the 
insurer,  is  based  upon  the  expectation  of  life  of  the  insured,  and 
on  the  rate  of  interest  or  net  profit  the  insurer  may  be  able  to 
make  by  investing  the  premium. 

449.  By  expectation  of  life  is  meant  the  average  number 
of  years  of  life  that  remains  to  any  individual  of  a  given  age, 
as  determined  by  the  rates  of  mortality. 

450.  The  Carlisle  Table  of  the  Expectation  of  Life,  which 
is  in  general  use  in  England,  has  also  been  taken  as  a  guide 
by  some  American  companies  in  fixing  their  rates  of  insurance. 

Other  companies  have  been  guided  in  fixing  their  premiums 
by  a  table  prepared  by  Dr.  Wigglesworth,  with  a  special  refer- 
ence to  mortality  in  this  country,  and  which  the  Supreme 
Court  of  Massachusetts  has  adopted  as  a  rule  in  estimating 
the  value  of  life  estates. 

451 .  The  expectation  of  life,  according  to  the  Carlisle  Table 

N      29 


338 


INSURANCE. 


and  according  to  that  prepared  by  Dr.  Wigglesworth,  is  shown 
in  the  following 

Table. 


4a 

5     d 

4d 

|  i 

.1   |d 

5     d 

•o  o 

t-)        o 

•^  ° 

•5  © 

u       0 

•2  °     ^       0 

6b 

< 

72  x 

CO  £}  -*3 

•  to 
<5 

©■g 

as. 
73  x 

0    •  -S 

GO  _Q  ■+* 

2    08     g 

.5?      « 

•      eh'-S 

§D           of       j 

<    §  &  i 

0    ."3 

CO    rt  +> 

I 

Hi       >   0  ■£ 

gfl 

*  ■* 

a» 

£     W 

48 

aw 

e  « 

72 

*W 

£     W 

0 

38.72 

28.15 

M 

38.59 

32.70 

22.80 

22.27 

8.16 

9.14 

1 

44.68 

36.78 

25 

37.86 

32.33 

49 

21.81 

21.72 

73 

7.72 

8.69 

2 

47.55 

38.74 

26 

37.14 

31.93 

50 

21.11 

21.17 

74 

7.33 

8.25 

3 

49.82 

40.01 

27 

36.41 

31.50 

51 

20.39 

20.61 

75 

7.01 

7.83 

4 

50.76 

40.73 

28 

35.69 

31.08 

69 

19.68 

20.05 

76 

669 

7.40 

5 

61.25 

40.88 

29 

35.00 

30.66 

53 

18.97 

19.49 

77 

6.40 

6.99 

G 

51.17 

40.69 

30 

34.34 

30.25 

54 

18.28 

18.92 

78 

6.12 

6.59 

7 

50.80 

40.47 

31 

33.68 

29.83 

17.58 

18.35 

79 

5.80 

6.21 

8 

50.24 

40.14 

32 

33.03 

29.43 

66 

16.89 

17.78 

80 

5.51 

5.85 

9 

49.57 

39.72 

33 

32.36 

29.02 

57 

16.21 

17.20 

81 

5.21 

5.50 

10 

48.82 

39.23 

34 

31.68 

28.62 

58 

15.55 

16.63 

82 

4.93 

5.16 

11 

48.04 

38.64 

35 

31.00  ' 

28.22 

59 

14.92 

16.04 

83 

4.65 

4.87 

12 

47.27 

38.02 

8fl 

30.32 

27.78 

60 

14.34 

15.45 

84 

4.39 

4.66 

13 

46.51 

37.41 

37 

29.64 

27.34 

61 

13.82 

14.86 

85 

4.12 

4.57 

14 

45.75 

36.79 

38 

28.96 

26.91 

62 

13.31 

14.26 

86 

3.90 

4.21 

15 

45.00 

36.17 

39 

28.28 

26.47 

63 

12.81 

;  13.66 

B7 

3.71 

3.90 

16 

44.27 

35.76 

40 

27.61 

26.04 

M 

12.30 

13.05 

88 

3.59 

3.67 

17 

43.57 

35.37 

41 

26.97 

25.61 

65 

11.79 

:  12.43 

89 

3.47 

3.56 

18 

42  87 

34.98 

42 

26.34 

25.19. 

66 

11.27 

11.96 

3.28 

3.73 

19 

42.17 

34.59 

43 

25.71 

24.77' 

67 

10.75 

11.48 

M 

3.26 

3.32 

20 

41.46 

34.22 

!  44 

25.09 

24.35 

68 

10.23 

;  11.01 

3.37 

3.12 

21 

40.75 

33.84 

48 

24.46 

23.92 

69 

9.70 

:  10.50 

1  93 

3.48 

!  2.40 

22 

40.04 

33.46 

46 

23.82 

23.37 

70 

9.18 

10.06 

M 

3.53 

j  1.98 

23 

39.31 

33.08 

1  47 

23.17 

22.88 

71 

8.65 

I    9.60 

1  95 

3.53 

1.62 

Against  any  age  given  in  the  table  may  be  found  the  expec- 
tation of  age  corresponding  to  it 

452.  The  transactions  of  life  insurance  companies  extend- 
ing as  they  do  over  a  term  of  years,  the  value  of  money  and 
the  average  rates  of  interest,  no  less  than  the  expectation  of  life, 
have  an  important  influence  in  fixing  their  rates  of  insurance. 

453.  The  premiums  of  life  insurance  are  generally  reck- 
oned at  a  certain  sum  on  $  100,  payable  annually  in  advance. 

The  rates  of  annual  premium  for  insuring  a  healthy  life  for 
one  year,  for  seven  years,  or  for  the  whole  period  of  life,  in  the 
sum  of  $  100,  by  the  Massachusetts  Hospital  Life  Insurance 
Company,  of  Boston,  and  by  the  Girard  Life  Insurance  An- 
nuity and  Trust  Company,  of  Philadelphia,  are  given  in  the 
following 


INSURANCE. 


339 


Table. 


Mass.  Hospital  Life  Insurance 

The  Girard  Life  Insurance 

Age  next 

Company. 

Company. 

Age  next 

Birthday. 

Birthday. 

1  year. 

7  years. 

For  Life. 

1  year. 

7  years. 

For  Life. 

15 

783~ 

.85 

1.44 

.88 

1.56 

15 

16 

.84 

.86 

1.47 

.84 

.90 

1.60 

16 

17 

.85 

.87 

1.51 

.86 

.91 

1.65 

17 

18 

.86 

.88 

1.54 

.89 

.92 

1.69 

18 

19 

.87 

.90 

1.58 

.90 

.94 

1.73 

19 

20 

.88 

.91 

1.62 

.91 

.95 

1.77 

20 

21 

.89 

.92 

1.66 

.92 

.97 

1.82 

21 

22 

.90 

.93 

1.70 

.94 

.99 

1.88 

22 

23 

.91 

.95 

1.74 

.97 

1.03 

1.93 

23 

24 

.92 

.96 

1.79 

.99 

1.07 

1.98 

24 

25 

.93 

.98 

1.84 

1.00 

1.12 

2.04 

25 

26 

.95 

.99 

1.89 

1.07 

1.17 

2.11 

26 

27 

.96 

1.01 

1.94 

1.12 

1.23 

2.17 

27 

28 

.98 

1.03 

2.00 

1.20 

1.28 

2.24 

28 

29 

.99 

1.05 

2.06 

1.28 

1.35 

2.31 

29 

30  ' 

1.01 

1.07 

2.12 

1.31 

1.36 

2.36 

30 

31 

1.03 

1.09 

2.18 

1.32 

1.42 

2.43 

31 

32 

1.05 

1.11 

2.25 

1.33 

1.46 

2.50 

32 

33 

1.07 

1.14 

2.32 

1.34 

1.48 

2.57 

33 

34 

1.09 

1.16 

2.40 

1.35 

1.50 

2.64 

34 

35 

1.11 

1.19 

2.48 

1.36 

1.53 

2.75 

35 

36 

1.14 

1.21 

2.56 

1.39 

1.57 

2.81 

36 

37 

1.16 

1.24 

2.65 

1.43 

1.63 

2.90 

37 

38 

1.19 

1.28 

2.75 

1.48 

1.70 

3.05 

38 

39 

1.22 

1.31 

2.85 

1.57 

1.76 

3.11 

39 

40 

1.24 

1.36 

2.95 

1.69 

1.83 

3.20 

40 

41 

1.27 

1.41 

3.07 

1.78 

1.88 

3.31 

41 

42 

1.31 

1.47 

3.19 

1.85 

1.89 

3.40 

42 

43 

1.35 

1.54 

3.32 

1.89 

1.92 

3.51 

43 

44 

1.40 

1.62 

3.45 

1.90 

1.94 

3.63 

44 

45 

1.47 

1.71 

3.60 

1.91 

1.96 

3.73 

45 

46 

1.54 

1.80 

3.75 

1.92 

1.98 

3.87 

46 

47 

1.62 

1.90 

3.92 

1.93 

1.99 

4.01 

47* 

48 

1.71 

2.02 

4.09 

1.94 

2.02 

4.17 

.    48 

49 

1.81 

2.14 

4.27 

1.95 

2.04 

4.49 

49 

50 

1.91 

2.28 

4.46 

1.96 

2.09 

4.60 

50 

51 

2.03 

2.42 

4.67 

1.97 

2.20 

4.75 

51 

52 

2.15 

2.59 

4.89 

2.02 

2.37 

4  90 

52 

53 

2.29 

2.76 

5.12 

2.10 

2.59 

5.24 

53 

54 

2.44 

2.95 

5.36 

2.18 

2.89 

5.49 

54 

55 

2.60 

3.15 

5.62 

2.32 

3.21 

5.78 

55 

56 

2.78 

3.38 

5.89 

2.47 

3.56 

6.05 

56 

57 

2.96 

3.62 

6.19 

2.70 

4.20 

6.27 

57 

58 

3.17 

3.87 

6.50 

3.14 

4.31 

•  6.50 

58 

59 

3.39 

4.17 

6.83 

3.67 

4.63 

6.75 

59 

60 

3.64 

4.50 

7.18 

J      4.35 

4.91 

7.00 

60 

According  to  the  table,  a  healthy  man  who  is  42  years  old 
next  birthday,  by  paying  the  Massachusetts  Hospital  Insurance 
Company  $  1.31,  would  secure  to  his  family,  heirs,  or  to  whom- 


340  INSURANCE. 

soever  he  desires,  $  100,  should  he  die  in  one  year,  and  in  the 
same  proportion  for  a  larger  sum.  And  if  he  would  obtain  a 
life  insurance  of  the  Girard  Life  Insurance  Annuity  and  Trust 
Company,  he  must  pay  annually  $  3.40.  The  New  York  Life 
Insurance  Company  has  a  schedule  of  rates  like  that  of  the 
Girard  Company,  given  in  the  table. 

454.  To  compute  the  premium  of  life  insurance  for  any 
given  amount. 

Ex.  1.  What  premium  will  the  Massachusetts  Hospital  Life 
Insurance  Company  require  for  the  insurance  of  a  life  one  year 
for  $  1728,  the  person  being  thirty  years  of  age  next  birth- 
day? Ans.  $17.45. 

operation.  By  the  table  we  find  the 

$1728  X  -0  101  =$17.4  5.     premium  on  $100  to  be  $1.01. 

Therefore  ${'-%-$  =  $  .0101 
=  the  premium  on  Si  for  one  year,  and  $  1728  X  .0101  =  the 
premium  on  $  1728  for  the  same  time.     Hence, 

Compute  the  premium  on  the  sum  to  be  insured  at  a  rate  propor- 
tionate to  the  given  premium  on  $  100. 

Examples. 

2.  What  amount  of  premium  must  S.  C.  Kendall  pay  an- 
nually to  the  Massachusetts  Hospital  Life  Insurance  Company, 
to  effect  an  insurance  on  his  life  for  7  years  for  $  8000,  his 
age  being  33  years  ?  Ans.  $  91.20. 

3.  Robert  Vaux,  60  years  of  age,  wishes  to  engage  in  a  very 
profitable  speculation;  and  being  obliged  to  borrow  the  neces- 
sary funds,  he  effects  an  insurance  on  his  life  for  7  years,  for 
$  78000,  at  the  office  of  the  Girard  Life  Insurance  Company. 
Required  the  amount  of  the  annual  premium.    Ans.  $  3829.80. 

4.  What  will  be  the  yearly  premium  for  insuring  a  person's 
life,  who  is  15  years  old,  for  $  2000  for  7  years,  at  the  New 
York  Life  Insurance  Company  ? 

5.  A  gentleman  45  years  of  age,  being  bound  on  a  long  and 
dangerous  voyage,  and  wishing  to  secure  a  competence  for  his 
family,  obtains  an  insurance  for  life  of  the  Girard  Life  Insur- 
ance Company,  for  $  12000.  By  an  act  of  Providence  he  dies 
in  the  third  year.     What  is  the  net  gain  to  his  family  ? 

Ans.  $  10657.20. 


CUSTOM-HOUSE   BUSINESS.  341 

6.  Richard  Sears,  50  years  old,  effects  an  insurance  for  life 
for  $  5000,  for  which  he  pays  an  annual  premium  of  $  4.60  on 
each  $  100  insured.  If  he  should  die  at  the  age  of  80  years, 
how  much  less  will  be  the  amount  of  insurance  than  the  pay- 
ments, allowing  the  latter  to  be  without  interest?   Ans.  $  1900. 

7.  A  gentleman,  56  years  old,  gets  his  life  insured  for  $  4000, 
at  the  office  of  the  Kentucky  Mutual  Life  Insurance  Company, 
by  paying  an  annual  premium  of  $5.20  on  each  $  100  insured ; 
and  dies  at  the  age  of  60  years.  Reckoning  interest  on  his 
payments  at  simple  interest,  what  is  gained  by  the  insur- 
ance? Ans.  $3043.20.. 

8.  Alexander  Murray,  28  years  of  age,  effects  an  insurance 
on  his  life  for  $  10000,  at  the  office  of  the  Massachusetts  Hos- 
pital Life  Insurance  Company.  If  the  company  loan  the  pre- 
mium at  6  per  cent,  compound  interest,  and  he  should  die  at 
the  age  of  40  years,  who  will  gain  by  the  insurance  ? 

Ans.  The  insured  gains  $  6626.01. 


CUSTOM-HOUSE  BUSINESS. 

455 •  Duties  or  customs  are  sums  of  money  required  by 
government  to  be  paid  on  imported  goods. 

Ports  of  entry  are  ports  into  which  merchandise  may  be  im- 
ported, or  from  which  it  may  be  exported. 

At  each  port  of  entry  is  an  establishment,  called  a  custom- 
house, at  which  certain  officers,  appointed  by  government,  at- 
tend to  the  collection  of  the  duties. 

456.     Duties  are  either  specific  or  ad  valorem. 

A  specific  duty  is  a  certain  sum  paid  on  a  ton,  hundred- 
weight, yard,  gallon,  &c,  without  regard  to  the  cost  of  the 
article. 

An  ad  valorem  duty  is  a  certain  percentage  paid  on  the  ac- 
tual cost  of  the  goods  in  the  country  from  which  they  are  im- 
ported. 

29* 


342  CUSTOM-HOUSE   BUSINESS. 

Note.  —  As  evidence  of  the  cost  value  of  merchandise  subject  to  duties,  the 
importer,  owner,  or  consignee  is  required  to  produce  an  invoice  or  manifest, 
if  one  has  been  received,  made  out  in  the  currency  of  the  place  or  country 
•whence  imported,  and  containing  a  true  statement  of  the  actual  cost  of  such 
goods  in  such  foreign  currency.  When  the  currency  of  a  place  has  a  depre- 
ciated falue  compared  with  that  of  the  United  States,  it  is  necessary  that  a 
consular  certificate  showing  the  rate  of  depreciation  should  be  attached  to  the 
invoice.  When,  however,  an  invoice  has  not  been  received,  the  fact  must  be 
testified  to  under  oath,  and  then  the  imported  articles  will  be  entered  at  an 
appraised  value. 

457 •  In  this  country,  under  the  present  tariff,  duties  are 
levied  on  the  ad  valorem  principle. 

458.  Allowances  are  deductions  required  to  be  made 
before  estimating  the  duties,  on  account  of  the  weight  of  the 
cask,  box,  bag,  &c.  in  which  an  article  is  imported,  or  on  ac- 
count of  breakage,  leakage,  waste,  or  other  damage. 

Tare  is  the  allowance  made  for  the  weight  of  the  cask,  box, 
&c.  containing  the  commodity. 

Draft  is  the  allowance  made  for  waste  in  the  weighing  of 
goods. 

Leakage  is  the  allowance  made  for  waste  on  liquids  imported 
in  casks. 

Gross  weight  is  the  weight  of  the  commodity  together  with 
the  cask,  box,  bag,  &c.  containing  it. 

Net  weight  is  what  remains  after  all  allowances  have  been 
made. 

459.  No  allowances  for  tare,  draft,  breakage,  &c.  are  appli- 
cable to  imports  subject  to  ad  valorem  duties,  except  actual 
tare,  or  weight  of  a  cask  or  box,  and  actual  drainage,  leak- 
age, or  damage.  The  collector  may  cause  these  to  be  ascer- 
tained, when  he  has  any  doubt  as  to  what  they  are. 

Note.  —  When  the  tariff  laws  of  the  United  States  required  the  collection 
of  specific  duties,  the  allowance  for  draft  was  on  1121b.,  lib.;  above  1121b.  and 
not  exceeding  2241b.,  21b. ;  above  2241b.  and  not  exceeding  3361b.,  31b. ;  above 
3361b.  and  not  exceeding  11201b.,  41b.;  above  11201b.  and  not  exceeding 
20161b.,  71b. ;  above  20161b.,  91b.  The  allowance  for  tare  was  deducted  after 
the  draft  had  been  deducted.  The  allowance  for  breakage  was  10  per  cent, 
on  all  beer  and  porter  in  bottles ;  and  5  per  cent,  on  all  other  liquors  imported 
in  bottles ;  and  a  dozen  bottles  of  common  size  were  estimated  to  contain  2| 
gallons.  The  allowance  for  leakage  was  2  per  cent,  on  liquors  imported  in 
casks.  In  making  the  allowances  for  tare  or  leakage,  a  fraction,  when  equal 
to,  or  greater  than,  one  half,  was  reckoned  1 ;  when  less,  it  was  omitted. 


CUSTOM-HOUSE    BUSINESS.  343 

Specific  duties  were  calculated  by  deducting  all  allowances  to  be  made  from  the 
given  quantity  of  merchandise,  and  multiplying  the  remainder  by  the  duty  on  a 
unit  of  the  given  quantity. 

460 •     To  calculate  ad  valorem  duties. 

Ex.  1.  What  is  the  duty  on  25651b.  of  sugar,  invoiced  at 
$  256.50,  at  24  per  cent,  ad  valorem  ?  Ans.  $  61.56. 

OPERATION. 

$  2  5  6.5  0  X  .2  4  =  $  61.56,  duty. 

Rule.  —  Find  the  percentage  on  the  invoiced  value  of  the  goods,  at 
the  given  rate  of  tariff,  and  the  result  will  be  the  ad  valorem  duty. 

!^ote  1.  —  Other  questions  in  duties  beside  those  requiring  the  finding  of 
the  amount  of  duty,  are  likewise  solved  by  some  one  of  the  rules  in  per- 
centage. 

Note  2.  —  In  custom-house  calculations  22401b.  are  considered  a  ton,  and 
1121b.  a  hundred-weight. 

Examples. 

2.  What  is  the  duty  at  8  per  cent,  on  an  importation  of 
books  invoiced  at  $  4350  ?  Ans.  $  348.00. 

3.  Required  the  duty  at  19  per  cent,  on  7890  pounds  of 
cordage,  invoiced  at  15  cents  per  pound.        Ans.  $  224.865. 

4.  What  is  the  duty  at  24  per  cent,  on  an  invoice  of  woollen 
goods,  which  cost  in  London  98 6£.,  the  pound  sterling  being 
valued  at  $  4.84?  Ans.  $  1145.3376. 

5.  $  112.50  duty  is  paid  on  an  importation  of  window-glass 
whose  invoice  value  was  $  750.  What  was  the  rate  per  cent, 
of  duty  ? 

6.  Robinson  &  Brother  of  New  York  have  imported  wines 
from  Havre,  invoiced  as  follows :  60  baskets  Champagne  at  70 
francs  per  basket;  36  baskets  port  at  35  francs  per  basket;  50 
casks  of  sherry,  each  31  gallons,  at  4  franca  per  gallon.  The 
allowance  for  breakage  on  the  wine  in  baskets  is  5  per  cent. ; 
and  the  actual  waste  of  that  in  casks  is  1  gallon  to  a  cask.  Re- 
quired the  duties  at  30  per  cent.,  allowing  the  value  of  a  franc 
to  be  18-ft  cents.  Ans.  $  624.2346. 

7.  Paid  $53.76  duties,  at  the  rate  of  8  per  cent,  on  60  casks 
of  raisins,  after  the  deduction  of  121b.  to  a  cask  for  tare.  Al- 
lowing the  gross  weight  of  each  cask  of  raisins  to  have  been 
1121b.,  what  was  their  invoice  value  per  pound? 

Ans.  11£  cents. 


344  CUSTOM-HOUSE   BUSINESS. 

8.  A  portion  of  the  cargo  of  the  ship  Cuba  from  Havana  was 
invoiced  as  follows :  40  hogsheads  of  molasses,  63  gallons  each, 
at  3  reals  plate  per  gallon ;  24  boxes  of  brown  sugar,  4001b. 
each,  at  1  real  vellon  per  pound;  260  boxes  of  oranges,  at  $2 
per  box;  and  410  boxes  of  cigars,  at  $7  per  box.  The  tare  on 
the  sugar  was  10  per  cent.,  and  the  leakage  of  the  molasses  2 
per  cent.  Allowing  the  value  of  a  real  plate  to  be  10  cents, 
and  a  real  vellon  5  cents,  and  the  rate  of  duties  on  the  molasses 
and  the  sugar  to  have  been  24  per  cent.,  on  the  oranges  8  per 
cent.,  and  on  the  cigars  30  per  cent.,  what  was  the  whole 
amount  of  duties  ? 

9.  270  tons  of  railroad  iron,  invoiced  at  $  50  per,  ton,  cost, 
when  the  duties  were  paid,  exclusive  of  other  charges,  $  1 6740. 
What  was  the  rate  per  cent,  of  duty  ?  Ans.  24  per  cent. 

10.  A  merchant  of  Baltimore  makes  an  importation  of  goods 
invoiced  at  $  20560.  On  goods  invoiced  at  $  3000  the  du- 
ties were  at  the  rate  of  4  per  cent. ;  on  goods  invoiced  at 
$  4200  the  duties  were  at  the  rate  of  8  per  cent. ;  on  goods  in- 
voiced at  $  2100  the  duties  were  at  the  rate  of  15  per  cent. ; 
goods  invoiced  at  $6000  were  free  of  duty;  and  on  the  re- 
mainder the  duties  were  at  the  rate  of  30  per  cent.  What  was 
the  whole  amount  of  the  duties  ? 

11.  Willard,  Fairbanks,  &  Co.,  of  Boston,  import  from  Liver- 
pool 10  pieces  of  Brussels  carpeting,  40  yards  each,  purchased 
at  5s.  per  yard,  duty  24  per  cent. ;  200  yards  of  hair-cloth,  at 
4s. per  yard, duty  19  per  cent.;  100  woollen  blankets,  at  2s.  6d., 
duty  15  per  cent. ;  and  shoe-lasting  to  the  cost  of  60£.,  duty  4 
per  cent.  Required  the  whole  amount  of  duty,  allowing  the 
value  of  the  pound  sterling  to  be  $  4.84.        Ans.  $  173.635. 

12.  A  merchant  imported  from  Bremen  32  pieces  of  linen  of 
32  yards  each,  on  which  he  paid  for  the  duties,  at  24  per  cent., 
$  122.88,  and  other  charges  to  the  amount  of  $  40.96.  What 
was  the  invoice  value  per  yard,  and  the  cost  per  yard  after  du- 
ties and  charges  were  paid  ? 

Ans.  Invoice  value  pes  yard,  $  0.50  ;  cost  per  yard,  $  0.66, 


COINS   AND    CURRENCIES. 


345 


COINS  AND  CURRENCIES. 

461.  Coins  are  pieces  of  metal  legally  stamped,  and  issued 
for  circulation  as  money. 

The  currency  of  a  state  or  country  is  its  money  or  circulating 
medium  of  trade. 

462.  The  former  currency  of  this  country  had  the  denomi- 
nations of  sterling  money,  viz.  pounds,  shillings,  and  pence.  On 
the  adoption  by  Congress,  in  1786,  of  the  present  decimal  cur- 
rency, with  the  dollar  as  its  unit,  there  were  in  circulation 
colonial  notes,  or  bills  of  credit,  which  had  depreciated  in  value. 
This  depreciation,  being  greater  in  some  States  than  in  others, 
gave  rise  to  the  difference  in  the  States  as  to  the  number  of 
shillings  equivalent  to  a  dollar,  as  shown  in  the  following 


Table, 


'  New  Eng.  States,"] 


$  1  in^ 


Virginia, 
Kentucky, 
Tennessee, 
fNew  York, 
.    J  Ohio, 
in1  Michigan, 

L  North  Carolina, 
"  Pennsylvania, 
New  Jersey, 
Delaware, 
^  Maryland, 
.    J  Georgia, 

\  South  Carolina, 


i  =====  6s.  =  i^£.,  called  New  Eng.  currency ; 
.j  of  which  l£.=  $3J-;  Is.  =  16 jets. 

==  8s.  ==■££.,  called  New  York  currency; 
of  which  l£.  ==  %  1\  ;  Is.  =  12£cts. 


1  in< 


i 


*  =  7s.  6d.  =| £.,  called  Pennsylvania  cur- 
rency ;  of  which  l£.==  $  2| ; Is.  ==  13  Jets. 


|  =4s.  8d.=^j-£.,  called  Georgia  currency ; 
J  of  which  l£.  =  $  4f :  Is.  ==  21f  cts. 

Note  1.  —  The  State  currencies  are  now  merely  nominal,  accounts  being  no- 
where kept  in  them.  Articles,  however,  are  sometimes  priced  in  them,  but 
much  less  often  now  than  formerly. 

Note  2.  —  $  1  equals  about  ^s-  =  hi^.  of  English  or  sterling  money;  and 
consequently  1£.  =  $4.84;  Is.  =  $0.24£.  Also  $  1  =  6s.  ==  £<£.  of  Nova 
Scotia,  New  Brunswick,  Newfoundland,  and  Canada,  called  Canada  currency; 
of  which  1«£.  =  $  4;  Is.  =  20  cents. 

463.     The  legal  tender  in  payment  of  debts  in  the  United 
States  is  gold  and  silver. 

Note.  —  The  gold  coins  of  the  United  States  of  coinage  prior  to  1834  are  a 
legal  tender  for  the  payment  of  all  debts,  at  the  rate  of  94g  cents  per  pemiy- 


346  COINS   AND   CURRENCIES. 

weight,  or  about  $  10.66  for  each  eagle;  and  all  the  gold  coins  of  a  subsequent 
coinage  are  a  legal  tender  of  payment  in  any  sums  whatever,  according  to 
their  nominal  values. 

Of  the  silver  coins  coined  prior  to  April  1, 1853,  the  three-cent  pieces  are  a 
legal  tender  of  payment  in  any  sums  of  thirty  cents  and  under,  and  the  dollars, 
half-dollars,  quarter-dollars,  dimes,  and  half-dimes  are  a  legar  tender  of  pay- 
ment in  any  sums  whatever,  at  their  nominal  value,  and  all  the  silver  coin  of  a 
subsequent  coinage  are  a  legal  tender  in  sums  not  exceeding  five  dollars. 

464.  Besides  the  coins  of  the  country,  foreign  coins,  whose 
value  has  been  fixed  by  law,  and  bank-notes,  redeemable  in 
specie,  pass  as  money.  However,  by  the  act  of  Congress  passed 
February,  1857,  all  former  acts  authorizing  the  currency  of  for- 
eign gold  or  silver  coin,  and  declaring  the  same  legal  tender 
in  payment  of  debts,  were  repealed. 

465.  The  intrinsic  value  of  foreign  coins  is  the  value  de- 
pending upon  the  weight  and  purity  of  the  metal  of  which  they 
are  made ;  their  legal  value  is  that  fixed  bylaw ;  their' commer- 
cial value  is  the  price  they  will  bring  in  the  market ;  and  their 
exchange  value  is  the  nominal  price  assigned  to  them  in  reck- 
oning exchange  between  one  country  and  another. 

466*  The  United  States  mint,  at  present,  purchases  stand- 
ard silver  at  $  1.22£  per  ounce,  and  fine  silver  at  $  1.36  per 
ounce. 

Note.  —  At  these  rates  of  purchase,  five-franc  pieces  yield  about  99  cents 
each;  Mexican  and  South  American  dollars,  106|  cents  each;  old  Spanish  dol- 
lars, 105  cents  each;  half-dollars  of  the  United  States  coined  before  1837, 
52£  cents  each;  half-dollars  of  the  United  States  coined  since  1837,  and 
previous  to  the  change  of  standard  in  1853,  52£  cents  each;  German  florins, 
414  cents  each;  Prussian  and  Hanoverian  thalers,  72  cents  each;  best  manu- 
factured American  plate,  from  120  to  122  cents  per  ounce;  and  genuine  British 
plate,  125  cents  per  ounce. 

467.  By  the  act  of  1843  the  gold  coins  of  Great  Britain,  if 
not  less  than  tttt  in  fineness,  were  rated  at  94T60-  cents  per  pen- 
nyweight, and  the  gold  coin  of  France  of  not  less  than  T8e9<$y  in 
fineness,  at  92T9J  cents  per  pennyweight.  By  a  previous  act, 
the  gold  coins  of  Portugal  and  Brazil  of  not  less  than  22  carats 
fine  were  rated  at  94T8IT  cents  per  pennyweight,  and  the  gold 
coins  of  Spain,  Mexico,  and  Colombia,  of  a  fineness  of  20  carats 
3y7^  grains,  at  the  rate  of  89T97  cents  per  pennyweight. 

Note.  —  The  relative  value  of  gold  and  silver  in  the  United  States,  at  pres- 


COINS   AND    CURRENCIES. 


347 


ent,  is  nearly  as  14£  to  1 ;  in  England,  as  14fQ95  to  1 ;  in  France  and  in  Russia,  as 
15  to  1 ;  in  Spain,  as  16  to  1 ;  in  China,  as  14£  to  1 ;  and  in  Portugal,  as  13^ 
tol. 

468.  The  value  of  certain  foreign  coins  and  currencies,  as 
fixed  by  law  in  the  collection  of  duties  at  the  United  States 
custom-houses,  is  shown  in  the  following 


Table. 


Pound  Ster.  of  G.  Britain,  $ 
Pound  Ster.  of  Br.  Prov.,^ 
Nova  Scotia,  N.  Bruns.,  > 
and  Newfoundland,        ) 
Dollar  of  Mexico,   Peru, 
Chili,  and  Cen.  Amer., 
Specie  Dollar  of  Sweden 

and  Norway, 
Specie  Dollar  of  Denmark, 
Rix  Dollar  of  Bremen, 
Thaler  of  Bremen  of  72 

grotes, 
Rix  Dollar,  or  Thaler  of 
Prussia    and    Northern 
States  of  Germany, 
Ruble,  silver,  of  Russia, 
Florin  of  the  Austrian  Em-" 
pire  and  city  of  Augs- 
burg, 
Florin  or  Guilder  of  Neth- 
erlands, 


4.84 
4.00 

1.00 

1.06 

1.05 

•  78f 

.71 

.69 

.75 


.481 


.40 


Florin  of  South  of  Germ.,  $  0.40 

Ounce  of  Sicily, 

Pagoda  of  India, 

Star  Pagoda  of  Madras, 

Tael  of  China, 

Millrea  of  Portugal, 

Millrea  of  Azores, 

Ducat  of  Naples, 

Sicca  Rupee  of  Bengal,  or  7 

of  Bombay,  £ 

Rupee  of  British  India, 
Mark  Banco  of  Hamburg, 
Franc  of  France  and  Belg., 
Livre  Tournois  of  France, 
Leghorn  Livre, 
Lira  of  Lombardo-Vene 

tian  Kingdom, 
Lira  of  Tuscany, 
Lira  of  Sardinia, 
Real  Plate  of  Spain, 
Real  Vellon  of  Spain, 


2.40 
1.94 
1.94 
1.48 
1.12 
.831 
.80 

.50 

.35 

.181 
.16 


;■• 


16 
.16 

.10 
,05 


REDUCTION  OF  CURRENCIES. 


469.  Reduction  of  Currencies  is  the  process  of  finding  the 
value  of  the  denominations  of  one  currency  in  the  denomina- 
tions of  another. 

470*  To  find  the  value  of  the  denominations  of  one  currency 
in  the  denominations  of  another,  when  the  values  of  a  unit  of 
each  are  known. 

Ex.  1.  What  is  the  value  of  18£.  4s.  6d.  of  the  New  Eng- 
land currency  in  United  States  money  ?  Ans.  $  60.75. 

operation.  We  reduce  the  shillings  and  pence 

18£.  4s.  6d.  =  18.225£.       to  the  decimal  of  a  pound,  and  an- 
18.225  -7-  -\  =  $  60.75.      nex  '*  to  tne  pounds.    We  then  di- 
vide the  sum  by  ^  because  6s.,  or 
a  dollar  in  this  currency,  is  ^  of  a  pound. 


348  COINS   AND   CURRENCIES. 

Ex.  2.     What  is  the  value  of  $  60.75  in  New  England  cur- 
rency ?  Ans.  18£.  4s.  6d. 

operation.  Since  6s-  >  or  a  dollaT  of  th[*  ?ur 

60.75  X  A  -  18.225£.      **■*■ 1S  A  of  a  Pou"d'  ?e  *?ultlP ^ 

1Q   00^  IQr      >|         *A  the    g1VCn    SUm    ^    the    fraCt1011    A» 

18.225£.  =±  18£.  4s.  6d.      and  find  the  value  of  tiie  decimal 

in  shillings  and  pence. 

Rule.  —  Multiply  or  divide,  as  the  case  may  require,  by  the  value  of 
the  unit  of  the  given  currency  expressed  in  United  States  money. 

Note.  —  When  one  currency  is  to  be  changed  to  another,  and  neither  of 
them  is  United  States  money,  the  value  of  one  of  them  may  be  found  in  United 
States  money,  and  the  value  of  that  result  be  found  in  the  other  currency. 

Examples. 

3.  Change  46£.  16s.  6d.  of  the  currency  of  New.  York  to 
United  States  money.  Ans.  $  117.06£. 

4.  Change  $  1032  to  the  currency  of  Pennsylvania. 

Ans.  387£. 

5.  Find  the  value  of  $  515.70  in  Canada  currency. 

Ans.  128£.  18s.  6d. 

6.  Change  160.50  francs  to  United  States  money. 

7.  Change  $  728.41  to  English  money. 

Ans.  150£.  9s.  llT62JTd. 

8.  Find  the  value  of  12£.  12s.  of  the  currency  of  Georgia 
in  United  States  money.  Ans.  $  54. 

9.  Find  the  value  of  128£.  18s.  6d.  of  Canada  in  United 
States  money.  Ans.  $515.70. 

10.  Find  the  value  of  740.45  rubles,  silver,  of  Russia,  in 
United  States  money.  Ans.  $  555.33J . 

11.  Find  the  value  of  46£.  16s.  6d.  of  New  York  currency 
in  the  currency  of  New  England.  Ans.  35£.  2s.  4^-d. 

12.  Change  151  millreas  of  Portugal  to  real  plate  of  Spain. 

13.  Find  the  value  of  1000  specie  dollars  of  Norway  in 
francs. 

14.  A  merchant  of  Quebec  bought  in  London  30  pieces  of 
broadcloth  of  30  yards  each,  at  1 5  shillings  per  yard ;  what  is 
the  amount  of  his  bill  in  Canada  currency?    Ans.  81 6£.  15s. 

15.  A  merchant  of  Prussia  bought  in  Naples  silks  to  the 
amount  of  410  ducats ;  what  was  the  amount  of  his  purchase 
in  thalers  ?  Ans.  475  fj-  thalers. 


EXCHANGE.  349 


EXCHANGE. 


471 .  Exchange,  in  commerce,  is  the  paying  or  receiving 
money  in  one  place  for  an  equivalent  sum  in  another,  by  means 
of  drafts,  or  bills  of  exchange. 

472.  A  Bill  of  Exchange  is  a  written  order,  to  some 
person  at  a  distance,  to  pay  a  certain  sum,  at  an  appointed 
time,  to  another  person,  or  to  his  order. 

The  maker  or  drawer  of  a  bill  is  the  person  who  signs  it. 

The  buyer,  taker,  or  remitter  of  a  bill  is  the  person  in  whose 
favor  it  is  drawn. 

The  drawee  of  a  bill  is  the  person  on  whom  it  is  drawn,  who 
is  also  called  the  acceptor,  after  he  has  accepted  it. 

The  indorser  of  a  bill  is  the  person  who  indorses  it. 

The  holder  or  possessor  of  a  bill  is  the  person  in  whose  legal 
possession  it  may  be  at  any  time. 

473  o  Most  mercantile  payments  are  made  in  bills  of  ex- 
change, since  it  is  generally  more  convenient  to  discharge  debts 
by  means  of  them  than  by  cash  remittances.  For  example, 
suppose  A,  of  Boston,  is  creditor  to  B,  of  Baltimore,  $100 ;  and 
C,  of  Boston,  is  debtor  to  D,  of  Baltimore,  $  100 ;  both  these 
debts  may  be  discharged  by  means  of  one  bill.  Thus,  A  draws 
for  this  sum  on  B,  and  sells  his  bill  to  C,  who  remits  it  to  D,  and 
the  latter  receives  the  amount,  when  due,  from  B.  Here,  by  a 
transfer  of  claims,  the  Boston  debtor  pays  the  Boston  creditor, 
and  the  Baltimore  debtor  the  Baltimore  creditor ;  and  no  money 
is  sent  from  one  place  to  the  other.  The  same  would  take  place 
if  D,  of  Baltimore,  drew  on  C,  of  Boston,  and  sold  his  bill  to  B, 
of  Baltimore,  who  should  send  it  to  A,  of  Boston ;  the  effect,  in 
either  case,  being  merely  a  transfer  of  debt  and  credit. 

Bills  of  exchange  pass  from  hand  to  hand,  until  due,  like  any 
other  circulating  medium. 

474.  The  terms  of  a  bill  vary  according  to  the  agreement 
between  parties,  or  the  custom  of  countries.  Some  bills  are 
drawn  at  sight ;  others,  at  a  certain  number  of  days,  or  months, 
after  sight  or  after  date  ;  and  some,  at  usance,  which  is  the  cus- 
tomary or  usual  term  between  different  places. 
N       30 


350  EXCHANGE. 

Days  of  Grace. 

Days  of  grace  are  a  certain  number  of  days  granted  the  ac- 
ceptor for  paying  the  bill,  after  the  term  of  a  bill  has  expired. 
The  usual  time  allowed  in  this  country  is  three  days. 

Note  1.  —  In  some  States  three  days  of  grace  are  allowed  on  all  bills  of  ex- 
change payable  at  sight,  or  at  a  future  day  certain ;  but  in  other  States  sight 
drafts  are  excepted,  as  in  New  York,  Pennsylvania,  New  Jersey,  Maryland, 
Virginia,  Missouri,  Illinois,  Michigan,  Connecticut,  Rhode  Island,  Delaware,  &c. 

No  days  of  grace  are  allowed  on  bills  written  payable  on  demand,  or  which 
have  no  time  of  payment  expressed  in  them. 

Note  2.  —  In  reckoning  when  a  bill,  payable  after  date,  becomes  due,  the 
day  on  which  it  is  dated  is  not  included;  and  if  it  be  a  bill  payable  after  sight, 
the  day  of  presentment  is  not  included.  When  the  term  is  expressed  in  months, 
calendar  months  are  understood ;  and  when  a  month  is  longer  than  the  preced- 
ing, it  is  a  rule  not  to  go  in  the  computation  into  a  third  month. 

Thus,  if  a  bill  be  dated  the  28th,  29th,  30th,  or  31st  of  January,  and  payable 
one  month  after  date,  the  term  expires  on  the  last  day  of  February,  to  which 
the  days  of  grace  must,  of  course,  be  added ;  and  therefore  the  bill  becomes 
due  on  the  3d  of  March. 

Indorsing  Bills. 

475.  An  indorsement  of  a  bill  is  the  act  by  which  the 
holder  of  it  transfers  his  right  to  another.  It  is  usually  made 
on  the  back  of  the  bill,  and  must  be  in  writing. 


- 


476.  Bills  payable  to  order  are  transferred  only  by  indorse- 
ment and  delivery,  but  bills  payable  to  bearer  are  transferred 
by  either  mode.  On  transfer  by  delivery,  the  person  making  it 
ceases  to  be  a  party  to  the  bill ;  but  on  a  transfer  by  indorse- 
ment, he  is  to  all  intents  and  purposes  chargeable  as  a  new 
drawer. 

477.  An  indorsement  may  take  place  any  time  after  the 
bill  is  issued,  even  after  the  day  of  payment  has  elapsed. 

Note.  —  An  indorsement  may  be  restrictive,  giving  authority  to  the  in- 
dorsee to  receive  the  money  for  the  indorser,  but  not  to  transfer  the  bill  to 
another.  The  indorsement  for  a  part  of  the  money  only  is  not  valid,  except 
with  regard  to  him  who  makes  it.  The  drawer  and  acceptor  are  not  bound  by 
it.  After  the  payment  of  a  part,  however,  a  bill  may  be  indorsed  over  for  the 
residue.  The  indorsement  is  said  to  be  in  blank,  when  the  indorser  simply 
writes  his  name  upon  the  back  of  the  bill  to  make  it  transferable  by  delivery ; 
and  is  said  to  be  special,  when  the  indorser  directs  the  money  to  be  paid  to  some 
particular  person,  or  to  his  order.  If  the  indorser  would  avoid  all  liability,  he 
must  qualify  his  indorsement  by  the  words,  "  without  recourse,"  or  by  others 
of  the  same  import. 


EXCHANGE.  351 

Accepting  Bills. 

478.  An  acceptance  is  an  engagement  to  pay  a  bill  accord- 
ing to  the  tenor  of  the  acceptance,  which  may  be  either  absolute 
or  qualified. 

479.  An  absolute  acceptance  is  an  engagement  to  pay  a  bill 
according  to  its  request,  which  is  commonly  done  by  the  drawee 
writing  his  name  at  the  bottom,  or  across  the  body  of  the  bill, 
with  the  word  accepted. 

480.  A  qualified  acceptance  is  when  a  bill  is  accepted  con- 
ditionally. But  the  holder  is  not  obliged  to  receive  a  conditional 
or  partial  acceptance.  He  may  act  as  if  an  acceptance  had  been 
entirely  refused. 

Note.  —  When  a  bill  is  drawn  for  the  account  of  a  third  person,  and  is  ac- 
cepted as  such,  and  he  fails  without  making  provision  for  its  payment,  the 
acceptor  must  discharge  the  bill,  and  can  have  no  recourse  against  the  drawer. 
When  a  holder,  at  his  own  risk,  takes  a  conditional  or  partial  acceptance,  he 
must  give  immediate  notice  to  all  other  parties  to  the  bill,  or  he  can  have  no 
resort  to  them  in  default  of  payment. 

All  bills  payable  at  sight,  or  at  a  day  certain,  or  on  demand,  should  be  pre- 
sented within  a  reasonable  time,  or  the  holder  may,  from  his  default,  be  the 
loser. 

Protesting  Bills. 

481  •  The  holder  of  a  bill,  when  acceptance  or  payment  has 
been  refused,  should  give  regular  and  immediate  notice  to  all 
the  parties  to  whom  he  intends  to  resort  for  payment ;  since,  if 
a  loss  should  be  incurred,  on  account  of  unnecessary  delay,  by 
the  failure  of  any  of  the  parties,  he  would  be  obliged  to  bear 
the  loss.  Such  a  notice  of  non-acceptance  or  non-payment, 
when  made  by  a  public  officer  called  a  notary  or  notary  public, 
or  by  any  other  legal  mode,  is  called  a  protest. 

482.  When  the  parties  to  a  bill,  which  the  drawee  has  failed 
to  accept  or  pay,  live  in  different  countries,  a  protest  is  indis- 
pensably necessary,  as  this  instrument  is  admitted  in  foreign 
countries  as  legal  proof  of  the  fact  of  the  refusal,  and  that  the 
holder  intends  to  recover  any  damages  which  he  may  sustain  in 
consequence. 

483.  In  case  of  non-payment,  when  the  parties  to  a  bill  live 
in  the  same  country,  it  is  not  necessary,  although  frequently 


352  .  EXCHANGE; 

practised,  that  there  should  be  a  regular  protest  by  a  public 
notary.  But  a  notice  simply  of  non-payment  is  sufficient  to  en- 
title the  holder  to  claim  interest. 

484.  The  damages  incurred  by  non-acceptance  and  non-pay- 
ment of  a  foreign  bill,  besides  interest  commencing  from  the 
day  of  demand,  consist  usually  of  the  exchange  or  re-exchange, 
commission,  and  postage,  together  with  the  expenses  of  protest 
and  interest.  The  damages  of  protested  bills  in  general,  how- 
ever, are  regulated  in  a  great  measure  by  the  local  laws  and 
usages  of  the  different  States  and  countries. 

Liabilities  of  the  Parties. 

485.  The  drawer,  acceptor,  and  each  and  every  indorser 
of  a  bill,  are  liable  to  the  payment  of  it ;  and  though  the  holder 
can  have  but  one  satisfaction,  yet,  till  such  satisfaction  is  ac- 
tually had,  he  may  sue  any  of  them,  or  all  of  them,  either  at 
the  same  time  or  in  succession,  and  obtain  judgment  against 
them  all,  till  satisfaction  be  made. 

486.  Nothing  will  discharge  an  indorser  from  his  engage- 
ment, but  the  absolute  payment  of  the  money ;  not  even  a  judg- 
ment recovered  against  the  drawer,  or  any  previous  indorser, 
or  any  execution  against  any  of  them,  unless  the  money  be  paid 
in  consequence. 

Note.  —  When  acceptance  is  refused,  and  the  bill  is  returned  by  protest,  an 
action  may  be  commenced  immediately  against  the  drawer,  though  the  regu- 
lar time  of  payment  be  not  arrived.  His  debt,  in  such  a  case,  is  considered  as 
contracted  the  moment  the  bill  is  drawn. 

In  order,  however,  to  make  the  indorsers  liable,  it  is  proper  that  the  holder 
should  present  the  bill  for  payment  on  the  day  it  becomes  due. 

Par  of  Exchange. 

487*  The  intrinsic  par  of  exchange  is  the  value  the  coins 
of  one  country  have,  when  compared  with  those  of  another,  with 
respect  both  to  weight  and  fineness. 

488.  The  commercial  par  of  exchange  is  the  value  the 
coins  of  one  country  sell  for  in  the  markets  of  another ;  and  is 
therefore  not  a  fixed,  but  a  variable  value. 


EXCHANGE.  353 

Course  of  Exchange. 

489.  The  course  of  exchange  is  the  variable  price  of  the 
money  of  one  country,  which  is  paid  for  a  fixed  sum  of  money 
of  another  country. 

490.  The  fluctuations  of  exchange  are  occasioned  by  vari- 
ous circumstances,  both  political  and  commercial,  but  in  general 
bills  rise  or  fall  in  their  prices,  like  any  other  salable  articles, 
according  to  the  relation  existing  for  the  time  being  between 
the  demand  and  the  supply. 

491.  The  limits  within  which  the  fluctuations  of  exchange 
range,  correspond  with  the  cost  of  making  remittances  in  cash. 
Therefore,  in  the  time  of  peace,  exchange  seldom  remains  long 
unfavorable  to  any  country.  When  unfavorable,  it  has  a  ten- 
dency to  correct  itself,  by  giving  an  unusual  stimulus  to  expor- 
tation, and  by  throwing  obstacles  in  the  way  of  importation ; 
and  when  favorable,  it  produces  the  same  effect,  by  restricting 
exportation  and  facilitating  importation. 

INLAND  BILLS. 

492.  An  Inland  Bill  of  exchange,  or  draft,  is  one  of 
which  the  drawer  and  drawee  are  residents  of  different  parts  of 
the  same  country. 

Inland  bills  are  seldom  bought  or  sold  at  the  precise  sum 
specified  upon  their  face,  but,  according  to  the  course  of  ex- 
change, are  subject  to  a  discount,  or  command  a  premium. 

493.  To  compute  inland  exchange. 

Ex.  1.  What  is  the  value  of  the  following  bill  of  exchange 
or  draft,  at  ^  per  cent,  premium  ?  Ans.  $  2563.20. 

$  2560.  New  York,  April  14,  1857. 

At  sight,  pay  to  Dura  Wadsworth,  or  order,  two  thousand  five 
hundred  and  sixty  dollars,  value  received,  and  charge  the  same 
to  the  account  of  Cameron,  Bashford,  &  Co. 

To  Messrs,  Lawrence  &  Aspinwall,  Merchants,  Boston. 

OPERATION. 

$25  60  X  1.0  0  4=  $25  63.2  0. 
30* 


354  EXCHANGE. 

Rule.  —  Multiply  the  face  of  the  hill  or  draft  by  1  increased  by  the 
rate  per  cent,  of  premium,  or  by  1  decreased  by  the  rate  per  cent,  of  dis- 
count, expressed  decimally.  The  product  will  be  the  value  of  the  given 
bill  or  draft. 

Note.  —  When  there  is  interest  to  be  computed,  it  must  be  reckoned  on  the 
face  of  the  bill  or  draft.  When  other  than  the  value  or  cost  of  the  bill  or  draft 
is  to  be  found,  proceed  as  in  percentage. 

Examples. 

2.  What  is  the  value  of  the  following  bill,  or  draft,  at  J  of  1 
per  cent,  premium?  Ans.  $  1955.37. 

$  1950^.  Chicago,  August  3,  1857. 

Thirty  days  after  date,  please  pay  to  the  order  of  Robert  S. 
Davis  §  Co.,  one  thousand  nine  hundred  fifty  -j^f  dollars,  value 
received,  and  charge  the  same  to  our  account 

Keene  &  Lee. 
To  George  Reed,  Broker,  Boston. 

3.  What  is  the  cost  of  a  draft  on  Philadelphia,  at  £  per  cent 
premium,  the  face  being  $  2000  ?  Ans.  $  2010. 

4.  What  must  be  the  face  of  a  draft,  at  2  per  cent,  discount, 
to  cost  $  1744.40  ? 

5.  What  is  the  cost  of  a  60  days'  bill  on  Pittsburg  to  the 
amount  of  $  600,  at  1  per  cent,  discount,  and  interest  off  at  6 
per  cent.  ?  Ans.  $  587.70. 

6.  What  is  the  cost  of  a  30  days'  bill  on  Boston,  at  f  per 
cent,  premium,  and  interest  off  at  6  per  cent.,  the  face  of  the  bill 
being  $  9256.40  ?  Ans.  $  9240.20. 

7.  What  must  be  the  face  of  a  30  days'  bill  which  will  yield 
$  9240.20  when  sold  at  f  per  cent,  premium,  and  interest  off  at 
6  per  cent.  ? 

8.  A  15  days'  draft  yielded  $  1190.184  when  sold  at  1£  per 
cent,  discount,  and  interest  off  at  6  per  cent.  What  was  the 
face  of  the  draft  ?  Ans.  $  1212. 

FOREIGN  BILLS. 

494.  A  Foreign  Bill  of  exchange  is  one  of  which  the 
drawer  and  drawee  are  residents  of  different  countries. 

495.  Foreign  bills  are  usually  drawn  in  sets;  that  is,  at 


EXCHANGE. 


355 


the  same  time  there  are  drawn  two  or  more  bills  of  the  same 
tenor  and  date,  each  containing  a  condition  that  it  shall  con- 
tinue payable  only  while  the  others  remain  unpaid.  Each  bill 
of  a  set  is  remitted  in  a  different  manner,  and  when  one  of  the 
set  has  been  accepted  and  paid,  the  others  become  worthless. 

In  reckoning  the  value  of  foreign  bills  of  exchange,  an  ac- 
quaintance with  the  moneys  of  account  of  foreign  countries  is 
required. 

498.  The  moneys  of  account  of  the  principal  places  of  com- 
mercial importance,  with  the  par  value  of  the  unit,  expressed  in 
United  States  money,  are  shown  in  the  following 

Table. 


Cities  and  Countries. 


London,  Liverpool,  &c, 

Paris,  Havre,  &c, 

Amsterdam,  Hague,  &c., 

Bremen, 

Hamburg,  Lubec,  &c, 

Berlin,  Dantzic,     < 

Belgium, 

St.  Petersburg, 

Stockholm, 

Copenhagen, 

Vienna,  Trieste,  &c, 

Naples, 

Venice,  Milan,  &c, 

Florence,  Leghorn,  &c, 

Genoa,  Turin,  &c, 

Sicily, 

Portugal, 

Spain, 

Constantinople, 
British  India, 
Canton, 


Denominations  of  Money. 


12  pence  =  1  shilling;  20s.  =  1  pound  = 

100  centimes  =  1  franc  m 

100  cents  =  1  guilder  or  florin  = 

5  swares  =  1  grote ;  72gr.  =  1  rix  dollar  = 

12  pfennings  =  1  schilling;   16s.  =  1  mark 

banco  = 
12  pfennings  =  1  groschen;  30gr.=l  thaler= 
100  centimes  =  1  franc  = 
100  kopecks  =  1  ruble  = 
12  rundstycks  =  16  skillings ;   48s.  =  1  rix 

dollar  specie  = 
16  skillings  =  1  mark ;  6m.  =  1  rix  dollar  = 
60  kreutzers  =  1  florin  = 
10  grani  =  1  carlino ;  lOcar.  =  1  ducat  = 
100  centesimi  =  1  lira  = 
100  centesimi  =  1  lira  = 
100  centesimi  =  1  lira  = 
20  grani  =  1  taro ;  30  tari  =  1  ounce  = 
1000  reas  =  1  millrea  = 

J  34  maravedis  =  1  real  vellon  = 

(  68  maravedis  =  1  real  plate  = 
100  aspers  =  1  piaster  = 
12  pice  =  1  anna;  16  annas  =  1  rupee  = 
100  candarines  =  1  mace ;  10m.  =  1  tael  = 


Value. 


1.06 

1.05 

0.48£ 

0.80 

0.16 

0.16 

0.186 

2.40 

1.12 

0.05 

0.10 

0.05 

0.44£ 

1.48 


Note  1.  —  Grani  is  the  plural  of  grano,  carlini  of  carlino,  centesimi  of  cente- 
simo,  lire  of  lira,  tari  of  taro. 

Note  2.  —  The  moneys  of  account  in  Brazil  are  of  the  same  denominations 
as  in  Portugal ;  in  Mexico,  Central  America,  New  Granada,  Chili,  Venezuela, 
Bolivia,  Peru,  and  Buenos  Ayres,  accounts  are  kept  in  pesos,  or  dollars,  of  8 
reals  each ;  in  Cuba  accounts  are  kept  in  dollars  equal  8  reals  plate,  or  20  reals 
vellon,  equal  $  1 ;  in  British  American  Possessions  accounts  are  kept  in  the  de- 
nominations of  sterling  money,  but  of  a  depreciated  value  compared  with  the 
currency  of  the  same  denominations  in  England,  except  in  Canada,  where,  by 
an  act  of  the  Colonial  Parliament,  passed  in  1857,  the  decimal  currency  of  the 
United  States  has  been  adopted. 


356  EXCHANGE. 

497 1  The  exchange  value,  in  United  States  money,  of  the 
pound  sterling  of  Great  Britain  is  that  of  its  former  legal  value, 
or  $  4§-  =  $  4.44f ,  which  is  considerably  below  either  its  in- 
trinsic or  commercial  value.  The  commercial  value  is  gen- 
erally about  9  per  cent,  more  than  this  exchange,  or  nominal 
par  value. 

Thus,  nominal  par  value  being        =  $  4.44| 
Add  premium  at  9  per  cent.  =       .40 

The  commercial  par  value  will  be  =  $  4.84|. 

Therefore,  when  the  nominal  exchange  between  the  United 
States  and  Great  Britain  exceeds  9  per  cent,  premium,  it  is 
above  true  par ;  when  less,  it  is  below  true  par. 

498.  The  quotations  of  the  rates  of  exchange  of  the  United 
States  on  England  have  always  reference  to  the  old  par  value 
of  the  pound  sterling  in  United  States  money.  The  course  of 
exchange  of  the  United  States  on  France  is  so  many  francs  and 
centimes  payable  in  France  for  a  dollar  paid  here;  on  Holland 
it  is  so  many  cents  a  guilder  (of  Netherlands)  ;  on  Hamburg, 
so  many  cents  a  mark  banco ;  on  Bremen,  so  many  cents  a  rix 
dollar ;  and  so  on. 

Note  1.  —  The  course  of  exchange  of  London  on  France  is  so  many  francs 
and  centimes  payable  in  France  for  1£. ;  on  Amsterdam,  it  is  so  many  florins  for 
1£. ;  on  Hamburg,  so  many  schillings  for  1£. ;  on  Vienna  and  Trieste,  so  many 
florins  and  creutzers ;  on  Spain,  so  many  pence  paid  in  England  for  1  dollar  of 
plate  (  =  8|  reals  plate)  payable  in  Spain  ;  on  Lisbon,  so  many  pence  for  1 
millrea ;  on  Naples,  so  many  pence  for  1  ducat ;  and  so  on. 

Note  2.  —  The  value  of  1£.  sterling  from  7  to  lo|  premium  on  the  old  par 
value  of  $  4.44|,  is  shown  in  the  following 

Table. 


7  per  cent. 
7l         u 

premium, 

a 

7|         " 

a 

7| 

8           " 

u 

u 

84-        " 

C( 

4        " 

a 

S3              U 

u 

9    per  cent,  premium,  $4,844 


$4,756 
4.767 
4.778 
4.789 
4.80 
4.811 
4.822 
4.833 

Note  3.  —  In  this  country  the  quotations  of  foreign  exchanges  are  usually 
for  bills  payable  60  days  after  sight,  and  of  inland  or  domestic  exchanges,  for 
bills  payable  at  sight. 


91 

u 

u 

4.856 

9I 

a 

u 

4.867 

9f 

u 

u 

4.878 

10 

it 

a 

4.889 

ioi 

u 

u 

4.90 

10i 

n 

a 

4.911 

101 

a 

u 

4.922 

EXCHANGE.  357 

To  compute  foreign  exchange. 

Ex.  1.  What  should  be  paid  for  the  following  bill  at  9£  per 
cent,  premium  ?  Ans.  $  486.666. 

Exchange  for  £  100.  Philadelphia,  May  21,  1857. 

Sixty  days  after  sight  of  this,  my  first  Bill  of  Exchange, 
{second  and  third  of  the  same  date  and  tenor  unpaid,)  pay  to 
Lang  don  Shannon,  or  order,  one  hundred  pounds  sterling,  value 
received,  with  or  without  further  advice. 

William  P.  Brown. 
To  Messrs.  Peabody  &  Co.,  Bankers,  London. 

OPERATION. 

100  X  -V-  =  $444.44;  $444,444  X  1.095  =  $486,666. 

2.  What  is  the  cost  at  Amsterdam  of  a  bill  on  New  York  for 
$  340.67,  exchange  being  at  $  0.38  to  the  guilder  ? 


Ans.  896.50  guilders. 


OPERATION. 


340.67  ~  38  =  896.5  ==  896  guilders  50  cents. 

3.  What  must  be  paid  in  Boston  for  a  bill  on  Paris  for  3676 
francs,  exchange  being  5  francs  20  centimes  to  the  dollar  ? 

Ans.  $706.92T\. 

4.  What  is  the  value  of  a  bill  on  Hamburg  for  3000  marks 
10  schillings,  exchange  being  at  $  0.35  to  a  mark  banco  ? 

Ans.  $  1050.213-. 

5.  How  large  a  bill  can   be  purchased  on  Liverpool  for 
$81727.75,  exchange  being  at  9 J  per  cent,  premium? 

6.  Paid  $14400.12  for  a  bill  on  Havre  for  79000  francs; 
how  much  was  exchange  below  par  ?  Ans.  2  per  cent. 

7.  What  is  the  cost  of  a  draft  on  St.  Petersburg  for  5763 
rubles  75  kopecks,  exchange  being  at  74  cents  a  ruble  ? 

Ans.  $  4265.175. 

8.  What  must  the  face  of  a  bill  on  Lisbon  be  which  costs 
$  550.66,  exchange  being  at  $  1.10  a  millrea? 

Ans.  500  millreas  600  reas. 

9.  What  is  the  cost  at  Berlin  of  a  draft  on  Philadelphia  for 
$  10000,  exchange  being  at  $  0.68  a  thaler  ? 

Ans.  14705  thalers  26  groschen  5-J-|-f~  pfennings. 

10.  How  much  must  be  paid  in  Chicago  for  a  bill  on  Stock- 
holm amounting  to  400  specie  rix  dollars  12  skillings  ? 

Ans.  $  424.265. 


358  ARBITRATION   OP   EXCHANGES. 

11.  What  is  the  value  in  St.  Louis  of  a  draft  on  Dantzic  for 
300  thalers  20  groschen  10  pfennings,  exchange  being  at  $0.69 
a  thaler  ?  Ans.  $  207.47^. 

12.  What  must  be  the  face  of  a  draft  on  Calcutta  which  costs 
in  Boston  $5694,  when  exchange  is  at  $0.40  a  company 
rupee  ?  Ans.  14235  rupees. 

13.  How  much  must  be  paid  in  Naples  for  a  draft  on  Balti- 
more amounting  to  $  615.60,  the  value  of  a  ducat  in  exchange 
being  $0.80?  Ans.  769  ducats  5  carlini. 

14.  There  was  paid  in  New  Orleans  $7300  for  1500£. 
draft  on  Liverpool ;  at  what  per  cent,  of  premium  was  it 
purchased  ?  Ans.  9  J  per  cent. 

15.  What  must  be  the  face  of  a  draft  on  Paris  that  can  be 
bought  in  London  for  868£.  17s.  6d.,  exchange  at  23  francs  60 
centimes  a  pound  sterling  ?     Ans.  20505  francs  45  centimes. 

16.  How  much  must  be  paid  in  Genoa  for  a  bill  on  New 
York  whose  face  is  $2640,  when  exchange  is  at  $0.18  a 
lire  ?  Ans.  14666  lire  (j$%  centesimi. 

17.  When  a  bill  on  Paris  for  88128  francs  costs  $  17280,  at 
what  per  cent  is  the  rate  of  exchange  below  par  ? 

18.  Robert  Anderson,  of  Cincinnati,  has  consigned  a  cargo 
of  pork,  valued  at  17000£.,  to  Richard  Arkwright  &  Son,  Liv- 
erpool. Robert  S.  Davis  &  Co.,  being  about  to  import  an  in- 
voice of  books,  have  purchased  of  Anderson  a  bill  of  exchange, 
at  8£  per  cent,  premium,  for  the  value  of  the  said  cargo.  What 
should  they  pay  for  the  bill  ?  Ans.  $  81977.777. 


ARBITRATION  OF  EXCHANGES. 

500.  Arbitration  of  Exchanges  is  the  process  of  finding 
the  proportional  exchange  of  two  places  by  means  of  one  or 
more  given  intermediate  exchanges. 

501*  Exchange  effected  thus,  through  one  or  more  inter- 
mediate exchanges,  is  called  circular  exchange. 

The  exchange  when  made  through  a  single  intervening  ex- 


ARBITRATION    OF   EXCHANGES. 


359 


change  is  called  simple  arbitration,  and  when  through  two  or 
more  intervening  exchanges  is  called  compound  arbitration. 

502.  Since  the  actual  course  or  rate  of  exchange  between 
any  two  places  is  almost  always,  from  various  circumstances, 
different  from  the  arbitrated  course,  the  object  of  arbitration  is 
to  enable  an  individual  in  one  place  to  ascertain  whether  he  can 
most  advantageously  draw  and  remit  directly  between  his  own 
place  and  another,  or  circuitously  through  other  places. 

503.  Exchange  of  merchandise,  and  the  different  weights 
and  measures  of  different  countries,  may  be  arbitrated  in  the 
same  manner  as  bills  of  exchange  and  currencies. 

504.  The  value,  in  the  standards  of  the  United  States,  of 
the  principal  weights  and  measures  of  the  most  important  com- 
mercial places,  is  shown  in  the  following 

Table. 


England. 

Value. 

Avoirdupois  pound, 
Old  wine  gallon, 

lib. 

lgal. 

Imperial  gallon, 
Old  ale  gallon, 

1.20gal. 

1.22gal. 

Old  Winchester  bushel, 

lbu. 

Imp.  corn  bu.  (=  8  imp.  gal.),  1.03bu. 

Quarter  of  grain  (=8  imp. 

bu.),  8.25bu. 

Imperial  yard, 

36in. 

France. 

Kilogramme, 

2.201b. 

Quintal  =  lOOkil., 

220.541b. 

Millier  =  lOOOkil., 

2205.481b. 

Litre, 

2.11pt. 

Velt, 

2gal. 

Decalitre  =  10  litre, 

2.64gal. 

Hectolitre  =  100  litre, 

26.42gal. 

Hectolitre  =  100  litre, 

2.84bu. 

Metre, 

39.36in. 

Holland  and  Belgium. 

Pond, 

1.081b. 

Fr.  kilogramme, 

2.201b. 

Last,  marine,  =  2000  p. 

44101b. 

Vat  =  lOOkan  =  1  hectol.  Fr.,  26.42s 

Ahm  of  wine, 

41sral. 

Mudde=100  kop=l  hectolitre,  2.84bu. 

Last  of  grain, 

85.25bu. 

Amsterdam  ell, 

27in. 

Hamburg. 

Pound, 

1.061b. 

100  pounds, 

106.81b. 

Ahm  of  wine, 

38.25gal. 

Value. 

Fuder  =  6  ahms, 

229.5gal. 

Last  of  grain, 

89.64bu. 

Stock  =  l£  last, 

134.4bu. 

Brabant  ell, 

27.58in. 

Denmark  and  Norway. 

Pound, 

1.101b. 

Centner  =  100  pounds, 

110.281b. 

Viertel  of  wine, 

2.04gal. 

Anker  of  wine, 

lOgal. 

Ahm  =  4  ankers, 

40gal. 

Fuder  of  wine, 

237.16gal. 

Toende  or  barrel  of  grain, 

3.95bu. 

Last  =  12  toende 

47.50bu. 

Danish  ell, 

24.66in. 

Sweden. 

Pound, 

.931b. 

Pound  of  iron, 

.751b. 

Anker  of  wine, 

10.35gal. 

Eimer  of  wine, 

20.75gal. 

Ahm  =  2  eimers, 

41.50gal. 

Pipe  =  3  ahms, 

124.25gal. 

Tun  or  barrel  of  grain, 

4.16bu. 

Ell, 

23.36in. 

Bremen. 

Pound, 

1.091b. 

Centner, 

1161b. 

Viertel  of  wine, 

1.93gal. 

Anker  =  5  viertels, 

9.65gal. 

Oxhoft  =  6  ankers, 

58gal. 

Scheffel  of  grain, 

2bu. 

Last  =  40  scheffels, 

80.70bu. 

Ell, 

22.75in. 

360 


ARBITRATION   OF   EXCHANGES. 


Trieste. 
Pound, 
100  pounds, 
Eimer  of  wine, 
Staro  of  grain, 
Ell  for  silks, 
Ell  for  woollens, 

Constantinople. 
Quintal, 

Alma  for  liquids, 
Kisloz  of  grain, 
Pik,  commercial, 


Value. 

1.231b. 

123.601b. 

15gal. 

234bu. 

25.20in. 

26.60in. 


124.451b. 
1.37gal. 
.94bu. 
27in. 


Cantaro  or  arroba  of  wine, 
Moyo  of  wine  =16  arrojm, 
Botta  =  38ar.  of  wine  =  3&& 

of  oil, 
Fanega  of  grain, 
Cahiz  =  12  fanegas, 
Vara  or  yard, 

Talue. 
4.25gal. 

68gal. 

ir. 

127.5gal. 

1.57bu. 

18.91bu. 

33.37in. 

Cuba. 

Quintal, 
Arroba  of  wine, 
Fanega  of  grain, 
Vara, 

101.751b. 
4.1gal. 
3bu. 
33.34in. 

Naples. 

Rottolo,  1.961b. 

Cantaro  grosso=100  rottolo,  196.501b. 
Cantaro  piccolo,  1061b. 


Calcutta. 

Maund,  74.661b. 

Bazaar  maund,  82.131b. 

Guz,  36in. 

Russia. 

Pound,  .901b. 

Pood  =  40  pounds,  361b. 

100  pounds,  90.261b. 

Wedro  of  wine,  3.25gal. 

Sorokovy  =  40  wedros,  130gal. 

Chetwert  of  grain,  5.95bu. 

Arsheen,  28in. 

Sashen,  7  ft. 

Prussia. 

Pound,  1.031b. 

100  pounds,  Dantzic,  103.31b. 

Quintal  =  110  pounds,  113.421b. 

Eimer  of  wine,  18.14gal. 

Ahm,  39.66gal. 

Scheffel  of  grain,  1.52bu. 

Last  of  grain,  91bu. 

Berlin  Ell,  25.5in. 

Prussian  Ell,  26.28in. 

Portugal. 

Pound  or  arratel,  1.011b. 

Arroba  =  22  arratels,  22.261b. 

Quintal  =  4  arrobas,  89.051b. 

100  pounds  or  arratels,  101.191b. 

Almude  of  wine,  4.37gal. 

Tonelado  =  52  almude,  227.25gal. 

Movo,  23.03bu. 

Vara,  43.20in. 

Spain. 

Pound,  1.011b. 

Arroba  =  25  pounds,  25.381b. 

Quintal  =  4  arrobas,  101.521b. 

Cantaro  or  arroba  of  oil,  3.75gal. 

Note.  —  The  weights  and  measures  of  Mexico,  Central  America,  and  of  the 
republics  of  South  America  are  the  same  generally  as  those  of  Spain ;  of  Brazil, 
the  same  as  those  of  Portugal;  of  the  British  North  American  Provinces,  the 
same,  in  general,  as  in  England ;  and  of  Hayti,  the  weights  are  the  same  as  in 

the  United  States,  except  about  8  per  cent,  heavier,  and  the  measures  the 
same  as  in  France. 


Salmaof  oil, 

42.7".<:al. 

Cairo  of  wine 

264gal. 

Carro  of  grain, 

52.20bu. 

Canna, 

83in. 

Sicily. 

Cantaro  grosso, 

192.501b. 

Cantaro  sottile, 

1751b. 

100  Sicilian  pounds, 

701b. 

Tonna, 

9.38gal. 

Salma  prr< 

9.48bu. 

Salma  generale, 

7.62bu. 

Couna  or  yard, 

38.40in. 

Genoa. 

Cantaro  grosso, 

76.871b. 

Cantaro  sottile, 

69.891b. 

Mt-z/.arola, 

39.25gal. 

Mina  of  grain, 

3.50bu. 

Canna  piCCOla, 

87.50in. 

Canna  grossa, 

116.70in. 

Venice. 

100  pounds,  pesso  grosso, 

105.181b. 

100  pounds,  pesso  sottile, 

66.421b. 

Miro  of  oil, 

4.02gal. 

Anfora  of  wine, 

1.37gal. 

Staja  of  grain, 

2.27bu. 

Moggio  =  4  staji, 

9.08bu. 

Braccio  for  silks, 

24.84in. 

Braccio  for  woollens, 

26.64in. 

China. 

Catty, 

1.331b. 

Pecul, 

133.331b. 

Covid, 

14.62in. 

ARBITRATION   OF   EXCHANGES.  861 

505 #     To  compute  arbitration  of  exchanges. 

Ex.  1.  When  the  exchange  between  New  York  and  Lon- 
don is  at  a  premium  of  9  per  cent.,  and  that  between  London 
and  Paris  25  francs  to  a  pound  sterling,  how  much  must  be 
paid  in  New  York  for  a  bill  on  Paris  for  1000  francs  ? 

operation.  Since  $  44  = 

$40       ±=         9£.  l£.ofthenomi- 

109£.  =  100£.  nal    par  of  ex- 

1  £.  =      2  5  fr.  change  (Article 

1  0  0  Ofr.  =  $  —  497),  $40=9£.; 

IQ  j0  and  109£.of  the 

yw         ^A\,         „  vw^w  same    value   = 

40   X  109  X  1   XJ0M  =  $193J7+  Ang.   100£.  at  9  per 

9  X  *00  X  ^  :\  cent*  Premium. 

We  wnte  $e  terms  of  equivalent  value  as  antecedent  and  consequent, 
and  proceed  as  in  conjoined  proportion  (Art.  341). 

Note.  —  When  it  is  required  to  find  which  of  several  routes  of  exchange  is 
the  most  advantageous,  the  rate  of  exchange  by  each  route  may  be  determined 
first,  and  the  results  then  compared. 

Examples. 

2.  When  exchange  at  Lisbon  on  Paris  is  at  the  rate  of  5 
francs  95  centimes  per  millrea,  and  at  Paris  on  the  United 
States  at  5  francs  20  centimes  per  dollar,  how  much  must  be 
paid  in  Lisbon  to  cancel  a  demand  in  New  York  for  $  3500  ? 

Ans.  3058  millreas  823T9Treas. 

3.  A  merchant  in  Boston  wishes  to  pay  2000£.  in  Liverpool. 
Exchange  on  Liverpool  he  finds  is  at  10  per  cent,  premium,  on 
Paris  5  francs  20  centimes  to  a  dollar,  and  on  Hamburg  35  cents 
to  a  mark  banco  ;  and  the  exchange  between  France  and  Eng- 
land at  the  same  time  is  24  francs  to  a  pound  sterling,  and  that 
of  Hamburg  on  England  13 J-  marks  banco  to  a  pound  sterling. 
Which  is  the  most  advantageous  course  of  remittance,  that  di- 
rect to  Liverpool,  or  that  through  Paris  or  Hamburg  ? 

4.  A  merchant  of  St.  Louis  wishes  to  pay  a  debt  of  $  5000 
in  New  York.  The  direct  exchange  is  1^-  per  cent,  in  favor  of 
New  York,  but  on  New  Orleans  it  is  £  per  cent,  premium ;  and 
between  New  Orleans  and  New  York  at  \  per  cent,  discount. 
How  much  would  he  save  by  the  circular  exchange  compared 
with  the  direct  ?      j  Ans."$  87.56£. 

N      31 


362  ARBITRATION   OF   EXCHANGES. 

5.  A  merchant  in  Boston  owes  a  debt  of  97  GO  thalers  in 
Bremen,  to  pay  which  he  purchases  a  bill  on  London,  at  a  pre- 
mium of  9  per  cent.,  and  remits  the  same  to  his  agent  in  Eng- 
land, on  whom  his  creditor  is  requested  to  draw.  If  the  ex- 
change between  London  and  Bremen  be  at  the  rate  of  3  1<1. 
sterling  per  thaler,  and  the  charges  for  brokerage  J  per 
cent.,  how  much  must  have  been  the  cost  of  the  bill  in  New 
York?  Ans.  $6731.74+. 

6.  When  exchange  between  New  Orleans  and  Hamburg  is 
at  34  cents  per  mark  banco,  and  between  Hamburg  and  St. 
Petersburg  is  2  marks  8  schillings  per  ruble,  how  much  must 
be  paid  in  St.  Petersburg  for  a  bill  on  New  Orleans  for 
$  650  ?  Ans.  764  rubles  70jf  kopecks. 

7.  "When  exchange  in  Philadelphia  on  Boston  is  at  J  per 
cent,  premium,  and  on  Chicago  at  2  per  cent,  discount,  if  the 
exchange  between  Chicago  and  Boston  is  at  par,  how  much 
better  is  the  circuitous  route  of  exchange  between  Philadelphia 
and  Boston  than  the  direct? 

8.  A  merchant,  about  to  import  broadcloth,  finds  he  can  ob- 
tain the  quality  desired  in  Amsterdam  at  8  guilders  per  Am- 
sterdam ell ;  in  Berlin,  at  3  thalers  15  groschen  per  Berlin  ell ; 
and  in  England,  at  15  shillings  per  yard.  Exchange  being  on 
Amsterdam  at  40  cents  per  guilder,  on  Berlin  at  66  cents  per 
thaler,  and  on  England  at  9  J  per  cent,  premium,  and  the  freight 
being  the  same  in  each  case,  from  which  place  can  he  make  the 
importation  to  the  best  advantage  ?  Ans.  Berlin. 

9.  When  exchange  between  Washington  and  London  is  at 
8  per  cent,  premium,  and  between  London  and  Paris  25.25 
francs  per  pound  sterling,  what  sum  in  Washington  is  equal  to 
7000  francs  in  Paris  ? 

10.  A  merchant  in  London  remits  to  Amsterdam  1000£.  at 
the  rate  of  18d.  per  guilder,  directing  his  correspondent  at 
Amsterdam  to  remit  the  same  to  Paris  at  2  francs  10  centimes 
per  guilder,  less  J  per  cent,  for  his  commission  ;  but  the  ex- 
change between  Amsterdam  and  Paris  happened  to  be,  at  the 
time^-the  order  was  received,  at  2  francs  20  centimes  per 
guilder.  The  merchant  at  London,  not  apprised  of  this,  drew 
upon  Paris  at  25  francs  per  pound  sterling.  Did  he  gain  or 
lose,  and  how  much  per  cent.  ?       Ans.  Gain,  16f  f  per  cent. 


ALLIGATION.  363 


ALLIGATION. 


505.  Alligation  is  a  process  employed  in  the  solution  of 
questions  relating  to  the  compounding  or  mixing  of  articles  of 
different  qualities  or  values. 

It  is  of  two  kinds  :  Alligation  Medial,  and  Alligation  Alter- 
nate, 

ALLIGATION  MEDIAL. 

506.  Alligation  Medial  is  the  process  of  finding  the 
mean  or  average  rate  of  a  mixture  composed  of  articles  of 
different  qualities  or  values,  the  quantity  and  rate  of  each 
being  given. 

507.  To  find  the  average  value  of  several  articles  mixed, 
the  quantity  and  rate  of, each  being  given. 

Ex.  1.  A  grocer  mixed  2cwt.  of  sugar  worth  $  9  per  cwt. 
with  lcwt.  worth  $  7  per  cwt.  and  2cwt.  worth  $  10  per  cwt. ; 
what  is  lcwt.  of  the  mixture  worth  ?  Ans.  $  9. 

$9  X  2=$18  Since  2cwt.  at  $  9  per  cwt.  is  worth 

7X1—         7  $18,  lcwt.  at  $  7  per  cwt.  is  worth  $  7, 

JO  v/  2  =      2  0  an(*  2cwt-  at  $  10  per  cwt.  is  worth  $  20 ; 

2cwt.  -)-  lcwt.  -f-  2cwt.  =  5cwt.  is  worth 

5  )     $45  $18-j-$7-f$20  =  $45;  and  lcwt.  is 

&  n     »  worth  as  many  dollars  as  45  contains  times 

$9  Ans-  5,or$9. 

Rule.  —  Find  the  value  of  each  of  the  articles,  and  divide  the  sum 
of  their  values  by  the  number  denoting  the  sum  of  the  articles.  The 
quotient  will  be  the  average  value  of  the  mixture. 

Examples. 

2.  If  19  bushels  of  wheat  at  $  1.00  per  bushel  should  be 
mixed  with  40  bushels  of  rye  at  $0.66  per  bushel,  and  11 
bushels  of  barley  at  $  0.50  per  bushel,  what  would  a  bushel  of 
the  mixture  be  worth  ?  Ans.  $  0.727|. 

3.  If  3  pounds  of  gold  of  22  carats  fine  be  mixed  with 
3  pounds  of  20  carats  fine,  what  is  the  fineness  of  the  mix- 
ture? Ans.  21  carats. 

4.  If  I  mix  20  pounds  of  tea  at  70  cents  per  pound  with  15 
pounds  at  60  cents  per  pound,  and  80  pounds  at  40  cents  per 
pound,  what  is  the  value  of  1  pound  of  this  mixture  ? 

Ans.  $  0.47£§. 


364  ALLIGATION. 

ALLIGATION  ALTEENATE. 

508.  Alligation  Alternate  is  a  process  of  finding  in 
what  ratio,  one  to  another,  articles  of  different  rates  of  quality 
or  value  must  be  taken,  to  compose  a  mixture  of  a  given  mean 
or  average  rate  of  quality  or  value. 

509.  To  find  the  proportional  quantities  of  the  articles  of 
different  rates  of  value  that  must  be  taken  to  compose  a  mix- 
ture of  a  given  mean  rate  of  value. 

Ex.  1.  A  merchant  has  spices,  some  at  18  cents  a  pound, 
some  at  24  cents,  some  at  48  cents,  and  some  at  GO  cents. 
How  much  of  each  sort  must  be  taken  that  the  mixture  may 
be  worth  40  cents  a  pound  ? 

Ans.  lib.  at  18c;  lib.  at  24c;  lib.  at  8c;  Ulb,  at  60c. 


FIRST    OPERATION. 


Given  mean,  , 
40  cents. 


*  lib.  at  18c,  gain  22c  \  =  38c.,  lft>.  at  18c.    =  18c 

lib.  at  24c,  gain  16c  j    gain-  lib.  at  21c    =  24c 


111),  at  48c,  loss  8c 
lib.  at  60c,  loss  20c. 


f  loss. 

-i  1 1  > .  at  60c,  loss  10c.  J  44Ib 


lib.  at  48c   =  48c 
38c.,  l£lb.  at  60c.  =  90c 


180c 
180c  -S-  4£  =  40c  per  lb. 

Compared  with  the  given  mean  value,  by  taking  lib.  at  18cts.  there 
is  a  gain  of  22cts.,  by  taking  lib.  at  24cts.  a  gain  of  16ets.,  by  taking 
lib.  at  48cts.  a  loss  of  8cts.,  and  by  taking  lib.  at  60cts.  a  loss  of 
20cts.  Now  it  is  evident  that  the  mixture,  to  be  of  the  mean  or 
average  value  given,  should  have  the  several  items  of  gain  and  loss 
in  the  aggregate  exactly  offset  one  another.  This  balance  we  effect 
by  taking  £lb.  more  of  the  spice  at  60cts. ;  and  thus  have  a  mixture 
of  the  required  average  value,  by  having  taken,  in  all,  lib.  at  18cts., 
lib.  at  24cts.,  lib.  at  48cts.,  and  1-Jlb.  at  60cts.  We  prove  the  cor- 
rectness of  this  result  by  dividing  the  value  of  the  whole  mixture  by 
the  number  of  pounds  taken. 

second  operation.  Having  arranged  in  a  column 

20        10  the  rates  of  the  articles,  with  the 

8  ^^  4  «  given  mean  on  the  left,  we  con- 
3  ccc.  nect  t0gether  terms  denoting  the 
j  ^  '  rate  of  the  articles,  so  that  a  rate 
less  than  the  given  mean  is  united 
with  one  that  is  greater.  We  then  proceed  to  find  what  quantity  of 
each  of  the  two  kinds  whose  rates  have  been  connected  can  be  taken, 
in  making  a  mixture,  so  that  what  shall  be  gained  on  the  one  kind 
shall  be  balanced  by  the  loss  on  the  other. 

By  taking  lib.  at  18cts.  the  gain  will  be  22cts. ;  hence  it  will  require 


ALLIGATION.  365 

?yb.  to  gain  let. ;  and  by  taking  lib.  at  60cts.  the  loss  will  be  20cts. ; 
hence  it  will  require  -^-lb.  to  lose  let.  Therefore,  the  gain  on  -^lb. 
at  18cts.  balances  the  loss  on  -^lb.  at  60cts.  The  proportions  at 
these  rates  are,  then,  -£%  and  -^  or  (by  reducing  to  a  common  denom- 
inator) ¥24°o  and  ¥2¥2q,  or  (by  omitting  the  denominators,  which  do  not 
affect  the  ratio)  20  and  22,  which  is  obviously  the  same  result  as 
would  be  obtained  by  placing  against  each  rate  the  difference  be- 
tween the  rate  with  which  it  is  connected  and  the  mean  rate.  In  like 
maimer  we  determine  the  quantity  that  may  be  taken  of  the  other 
two  articles,  whose  rates  are  connected  together. 

We  thus  find  that  there  may  be  taken  -^-lb.  at  18cts.,  -^lb.  at 
24cts.,  |lb.  at  48cts.,  and  -^lb.  at  60cts. ;  or,  20lb.  at  18cts.,  8lb.  at 
24cts.,  161b.  at  48cts.,  and  22lb.  at  60cts.  By  dividing  the  last  set 
by  2,  we  obtain  another  set  of  results,  and  by  multiplying  or  dividing 
any  of  these  results  others  may  be  found,  all  of  which  can  be  proved 
to  satisfy  the  conditions  of  the  question.  Hence,  examples  of  this 
kind  admit  of  an  indefinite  number  of  answers. 

Rule  1.  —  Take  a  unit  of  each  article  of  the  proposed  mixture,  and 
note  the  gain  or  loss  ;  and  then  take  such  additional  quantity  or  quan- 
tities of  the  articles  as  shall  equalize  the  gain  and  loss.     Or, 

Rule  2.  —  Write  the  rates  of  the  articles  in  a  column,  with  the  mean 
rate  on  the  left,  and  connect  the  rate  of  each  article  which  is  less  than 
the  given  mean  ivith  one  that  is  greater ;  the  difference  between  the  mean 
rate  and  that  of  each  of  the  articles,  written  opposite  to  the  rate  with 
which  it  is  connected,  will  denote  the  quantity  to  be  taken  of  the  article 
corresponding  to  that  rate. 

Note.  —  When  a  rate  has  more  than  one  rate  connected  with  it,  the  sum  of 
the  differences  written  against  it  will  denote  the  quantity  to  be  taken.  There 
will  be  as  many  different  answers  as  there  are  different  ways  of  connecting  the 
rates ;  and,  by  multiplying  and  dividing,  these  answers  may  be  varied  indefi- 
nitely. 

Examples. 

2.  How  much  barley  at  45  cents  a  bushel,  rye  at  75  cents, 
and  wrheat  at  $1.00,  must  be  mixed,  that  the  composition  may 
be  worth  80  cents  a  bushel  ? 

Ans.  1  bushel  of  rye,  1  of  barley,  and  2  of  wheat. 

3.  A  goldsmith  would  mix  gold  of  19  carats  fine  with  some 
of  15,  23,  and  24  carats  fine,  that  the  compound  may  be  20 
carats  fine.     What  quantity  of  each  must  he  take  ? 

Ans.  loz.  of  15  carats,  2oz.  of  19,  loz.  of  23,  and  loz.  of  24. 

4.  It  is  required  to  mix  several  sorts  of  wine  at  60  cents,  80 
cents,  and  $1.20,  with  water,  that  the  mixture  may  be  worth 
75  cents  per  gallon ;  how  much  of  each  sort  must  be  taken  ? 

Ans.  lgal.  of  water,  lgal.  of  60  cents,  9gal.  of  80  cents,  and 
lgal.  of  $1.20. 

31* 


366  ALLIGATION. 

510.  When  the  quantity  of  one  or  more  of  the  articles  com- 
posing a  mixture  of  a  given  mean  value  is  given,  to  find  the 
quantity  of  each  of  the  others. 

Ex.  1.  How  much  gold  of  15,  17,  and  22  carats  fine  must 
be  mixed  with  5  ounces  of  18  carats  fine,  so  that  the  compo- 
sition may  be  20  carats  fine  ? 

Ans.  loz.  at  15  carats,  loz.  at  17  carats,  9oz.  at  22  carats. 

operation.  By  taking  loz.   at  15  carats 

{loz.  at  15,  gain  5  ^  fine  there  is  a  gain  of  5  carats, 
loz.  at  17,  gain  3  >-  =  18  by  taking  loz.  at  17  carats  a 
5oz.  at  18,  gain  10  )  £*in  of  3  carats,  by  taking  5oz., 
loz.  at  22,  loss  2  )  the  ?iven  quantity,  at  18  carats, 
(  __  ^g  a  gain  of  10  carats,  and  by  tak- 

8oz.  at  22,  loss  16  )  inS  loz-  at  22  carats  a  loss  of.2 

carats;  and  to  balance  the  gain 

and  loss  we  take  8oz.  additional,  at  22  carats,  a  loss  of  16.     We  have 

then  for  the  result  loz.  at  15  carats,  loz.  at  17  carats,  and  9oz.  at  22 

carats. 

Rule.  —  Take  of  the  limited  artiele  or  articles  the  quantify  or  quan- 
tities given,  with  a  unit  of  each  of  the  other  articles  of  the  proposed 
mixture,  and  note  the  gain  or  loss ;  and  tfu  n  take,  if  required,  such  ad- 
ditional quantity  or  quantities  of  the  articles  not  limited,  as  shall  equalize 
the  gain  and  loss. 

Examples. 

2.  How  much  wine  at  $1.75  and  at  $  1.25  per  gallon  must 
be  mixed  with  20  gallons  of  water,  that  the  whole  may  be  sold 
at  $  1 .00  per  gallon  ?  Ans.  20  gallons  of  each. 

3.  How  much  wheat  at  $  2.00  per  bushel  and  at  $  1.80  per 
bushel  must  be  mixed  with  4  bushels  at  $  2.20  per  bushel  and 
10  bushels  at  $  1.70  per  bushel  to  make  a  mixture  worth  §  1.90 
per  bushel  ?       Ans.  9  bushels  at  $  2.00 ;  1  bushel  at  $  1.80. 

4.  How  many  pounds  of  sugar,  at  8,  14,  and  13  cents  a 
pound,  must  be  mixed  with  three  pounds  at  9^-  cents,  4  pounds 
at  10J-  cents,  and  6'  pounds  at  13^-  cents  a  pound,  so  that  the 
mixture  may  be  worth  12£  cents  a  pound  ? 

Ans.  lib.  at  8cts.;  5£lb.  at  13cts.;  and  91b.  at  14cts. 

5.  How  much  barley  at  45  cents  a  bushel  must  be  mixed 
with  10  bushels  of  oats  at  58  cents  a  bushel,  to  make  a  mixture 
worth  50  cents  a  bushel  ? 


ALLIGATION.  367 

511.  When  the  quantity  and  rate  of  a  mixture,  with  the 
rates  of  the  articles  composing  it,  are  given,  to  find  the  quantity 
of  each  article  which  is  not  limited. 

Ex.  1.  How  many  gallons  of  water  must  be  mixed  with 
wine  at  $1.50  a  gallon  so  as  to  make  a  mixture  of  100  gallons 
worth  $  1.20  a  gallon  ?  Ans.  20  gallons. 

operation.  Representing  the 

1  90  $  *Sal'  a*  0-00?  gam  1-20     1«20  rate  of  the  water  by 

*       (  lgal.  at  1.50,  loss     .30 )  0.00,  we  then  find, 

V  1  20  as  m  Art.  5^9,  the 

3gal.  at  1.50,  loss     .90)  quantity  required  of 

°  J  each  article,  in  com- 

1  -|-  4  =  5gal. ;  \  of  lOOgal.  =  20gal.  Ans.  posing  a  mixture  of 

the  given  mean,  to 
be  1  gallon  of  water  and  4  gallons  of  the  wine.  Therefore,  the  quan- 
tity of  water  is  to  the  whole  quantity  of  the  mixture  as  1  to  5. 
Hence,  in  a  mixture  of  100  gallons  at  the  mean  rate  given,  the  water 
must  be  \  of  100  gallons,  or  20  gallons. 

Rule.  —  Find  the  proportional  quantities  of  the  several  articles,  as 
in  Art.  509,  or  510,  as  though  the  quantity  of  the  mixture  were  not 
limited. 

Then  take  such  a  part  of  the  given  quantity  of  the  mixture,  as  each 
of  these  proportional  quantities  is  of  their  sum. 

Examples. 

2.  A  merchant  has  sugar  at  8  cents,  10  cents,  12  cents,  and 
20  cents  a  pound ;  with  these  he  would  fill  a  hogshead  that 
would  contain  200  pounds.  How  much  of  each  kind  must  he 
take,  so  that  the  mixture  may  be  worth  15  cents  a  pound  ? 

Ans.  33p>.  of  8,  10,  and  12cts.,  and  1001b.  of  20cts. 

3.  How  much  wheat  at  $  2.00  and  $  1.80  a  bushel  must  be 
mixed  with  4  bushels  at  $  2.20,  and  10  bushels  at  $  1.70,  so  as 
to  make  a  mixture  of  48  bushels,  worth  $  1.90  per  bushel? 

Ans.  21  bushels  $  2.00;  13  bushels  at  $  1.80. 

4.  How  much  gold  of  15,  17,  and  22  carats  fine  must  be 
mixed  with  five  ounces  of  18  carats  fine,  to  make  a  compo- 
sition of  5  pounds,  that  shall  be  20  carats  fine  ? 

5.  A  gentleman's  servant  having  been  ordered  to  purchase 
20  animals  for  $  20,  brought  home  sheep  at  $  4.00,  lambs  at 
$  0.50,  and  kids  at  $  0.25  each.  Required  the  number  of  each 
kind.  Ans.  3  sheep  ;  15  lambs  ;  and  2  kids. 


368  MISCELLANEOUS   EXAMPLES. 


MISCELLANEOUS   EXAMPLES. 

1.  A  manufacturer  employs  a  number  of  men  at  $  1.20,  and 
a  number  of  boys  at  $  0.80,  per  day ;  and  the  amount  of  the 
wages  of  the  whole  is  the  same  as  if  each  had  $0.9  7 J-  per  day. 
Required  the  number  of  men,  that  of  the  boys  being  9. 

Ans.  7  men. 

2.  What  is  the  value  of  5000  specie  rix  dollars  12  skillings 
of  Sweden  in  United  States  money  ?  Ans.  $  5300.265. 

3.  Exchange  between  New  Orleans  and  England  being  in 
New  Orleans  at  8  per  cent,  premium,  and  in  Liverpool  at  10 
per  cent,  premium,  if  L.  Sandford  of  Liverpool  owes  M.  Las- 
sale  of  New  Orleans  for  cotton  to  the  amount  of  1500£.  15s. 
sterling,  what  will  be  the  difference  between  Lassale  drawing 
or  Sandford  remitting  the  amount  ? 

4.  If  17  gallons  of  spirits  at  $1.26  per  gallon  be  mixed 
with  7  gallons  at  a  different  price,  and  20  per  cent,  be  made  by 
selling  the  mixture  at  $  1.56,  what  was  the  price  of  the  latter 
kind  per  gallon?  Ans.  $  1.39f  per  gallon. 

5.  What  is  the  value  of  100  ounces  20  tari  10  grani  of 
Naples  in  lire  and  centesimi  of  Leghorn  ? 

Ans.  1510  lire  25  centesimi. 

6.  If  20  United  States  gallons  equal  1  eimer  of  Sweden,  3 
eimers  of  Sweden  equal  4  eimers  of  Trieste,  24  eimers  of 
Trieste  equal  9  ahms  Danish,  and  33  ahms  Danish  equal  5 
carri  of  Naples,  which  will  cost  the  most  in  United  States 
money,  170  eimers  of  Trieste  of  wine  at  1  florin  45  kreutzers 
per  gallon,  or  12  carri  of  wine  at  1  ducat  of  Naples  a  gallon? 

7.  When  exchange  on  England  is  at  8  per  cent,  premium, 
and  freight  at  12d.  per  United  States  bushel,  how  much  can  be 
paid  per  bushel  for  wheat  in  Baltimore,  in  answering  an  order 
from  Liverpool  limited  to  60s.  per  imperial  quarter  ? 

Ans.  $  1.50T6T  per  bushel. 

8.  A  merchant  mixes  11  pounds  of  tea  with  5  pounds  of  an 
inferior  quality,  and  gains  16  per  cent,  by  selling  the  mixture 
at  87  cents  per  pound.  Allowing  that  a  pound  of  the  one  cost 
12  cents  more  than  a  pound  of  the  other,  what  was  the  cost  of 
each  kind  per  pound  ? 

Ans.  The  one  78£cts. ;  the  other  66Jets.  per  lb. 


INVOLUTION.  369 


EVOLUTION. 

512.  Involution  is  the  process  of  finding  the  powers  of 
quantities. 

A  power  of  a  number  or  quantity  is  the  product  of  that  num- 
ber multiplied  into  itself  one  or  more  times. 

513.  The  number  from  which  a  power  is  derived  is  called 
the  root  of  that  power. 

The  first  power  is  the  root,  or  the  number  involved. 
The  second  power  is  the  product  of  the  root  multiplied  by  it- 
self once,  or  used  twice  as  a  factor. 

The  third  power  is  the  root  used  three  times  as  a  factor;  &c. 

514.  The  index  or  exponent  of  a  power  is  a  small  figure 
written  at  the  right,  above  the  root,  indicating  the  number  of 
times  it  is  employed  as  a  factor.  Thus,  the  second  power  of  4 
is  written  42,  the  third  power  of  9  is  written  93,  and  the  fourth 
power  of  f-  is  written  (J)4. 

Note.  —  In  denoting  the  power  of  a  fraction,  the  fraction  is  included  in  a 
parenthesis,  in  order  that  the  exponent  may  be  regarded  as  applying  to  the 
whole  expression,  and  not  to  the  numerator  alone.  When  no  index  is  written, 
the  number  itself  is  to  be  considered  the  first  power.  The  second  power  is 
sometimes  called  the  square  of  a  number,  the  third  power  the  cube,  and  the 
fourth  power  the  biquadrate. 

515.  To  raise  a  number  to  any  required  power. 

2=2,  the  first  power  of  2,  is  written  2l  or  2. 

2X2=    4,  the  second  power  of  2,  is  written      22. 

2X2X2=    8,  the  third  power  of  2,     "        "  23. 

2  X  2  X  2  X  2  =  1 6,  the  fourth  power  of  2,  "        "  24. 

2X2X2X2X2  =  32,  the  fifth  power  of  2,    "        "  25. 

By  examining  the  several  powers  of  2  in  the  examples,  it  is  seen 
that  each  has  been  produced  by  taking  the  2  as  a  factor  as  many 
times  as  there  are  units  in  the  exponent  of  each  power  raised. 
Hence  the 

Ktjle.  — Multiply  the  given  number  into  itself,  till  it  has  been  used  as 
a  factor  as  many  times  as  there  are  units  in  the  exponent  of  the  power 
to  which  the  number  is  to  be  raised. 

Note  1.  —  The  number  of  multiplications  will  always  be  one  less  than  the 
number  of  units  in  the  exponent  of  the  power  to  be  raised,  since  in  the  first 


370  INVOLUTION. 

multiplication  the  root  is  used  twice;  once  by  being  taken  as  the  multiplicand, 
and  once  more  as  the  multiplier. 

Note  2.  — A  fraction  is  involved  by  involving  both  its  numerator  and  its  de- 
nominator. 

Examples. 

1.  What  is  the  3d  power  of  8  ?  Ans.  512. 

2.  What  is  the  5th  power  of  4  ?  Ans.  1024. 

3.  What  is  the  3d  power  of  J  ?  Ans.  §  J. 

4.  What  is  the  4th  power  of  2J  ?  Ans.  50$f . 

5.  What  is  the  5th  power  of  -J  ?  Ans.  ^j. 

6.  What  is  the  6th  power  of  5  ?  Ans.  15625. 

7.  What  is  the  6th  power  of  If  ?  Ans.  16jf  w  j; 

8.  What  is  the  value  of  710  ?  Ans.  282475249. 

9.  What  is  the  value  of  .0454  ?  Ans.  .000004100625. 

516.  To  raise  a  number  to  any  required  higher  power,  with- 
out producing  all  the  intermediate  powers. 

Ex.  1.     What  is  the  7th  power  of  5  ?  Ans.  78125. 

OPERATION. 
12  3  3+2        +        2=  7 

5,  2  5,  12  5;     125x25x25  =  7812  5. 

We  raise  the  5  to  the  2d  and  to  the  3d  power,  and  write  above 
each  power  its  exponent.  Then,  by  adding  the  exponent  2  to  itself, 
and  increasing  the  sum  by  the  exponent  3,  we  obtain  7,  a  number 
equal  to  the  exponent  of  the  required  power;  and  by  multiplying  25, 
the  power  belonging  to  the  exponent  2,  into  itself,  and  the  product 
thence  arising  by  125,  the  power  belonging  to  the  exponent  3,  we  ob- 
tain 78125,  the  required  7th  power.     Therefore, 

The  product  of  two  or  more  powers  of  the  same  number  is  that  power 
which  is  denoted  by  the  sum  of  their  exponents.     Hence,  the 

Rule.  —  Multiply  together  two  or  more  powers  of  the  given  number, 
the  sum  ofichose  exponents  is  equal  to  the  exponent  of  the  power  re- 
quired, and  the  product  will  be  that  power. 

Note.  —  When  the  number  to  be  involved  contains  a  decimal,  it  is  generally 
sufficient  to  retain  in  the  result  not  more  than  six  places  of  decimals;  and  the 
work  may  be  accordingly  contracted  as  in  the  multiplication  of  decimals 
(Art.  273). 

Examples. 

2.  What  is  the  7th  power  of  8  ?  Ans.  2097152. 

3.  What  is  the  9th  power  of  7  ?  Ans.  40353607. 

4.  What  is  the  10th  power  of  6  ? 


EVOLUTION.  371 

5.  What  is  the  5th  power  of  195  ?    Ans.  281950621875. 

6.  What  is  the  6th  power  of  §  ?  Ans.  T6^4F. 

7.  Required  the  2d  power  of  4698. 

8.  Required  the  2d  power  of  6031.  Ans.  36372961. 

9.  What  is  the  13th  power  of  7  ?        Ans.  96889010407. 

10.  What  is  the  12th  power  of  6  ? 

11.  What  is  the  15th  power  of  9  ?    Ans.  205891132094649. 

12.  What  is  the  4th  power  of  4.367  ?      Ans.  363.691179+. 

13.  Involve  the  following  numbers  to  the  powers  denoted  by 
their  respective  exponents  :  (2£  )5,  1.04'5,  and  (3f)4. 

Ans.  157 -ft8A;  1.800943+ ;  116i|-g-f 


EVOLUTION. 


517  •  Evolution,  or  the  extraction  of  roots,  is  the  process 
of  finding  the  roots  of  quantities.  It  is  the  reverse  of  involu- 
tion. 

518.  The  root  of  a  quantity  or  number  is  such  a  factor  as, 
being  multiplied  into  itself  a  certain  number  of  times,  will  pro- 
duce that  quantity  or  number. 

The  root  takes  the  name  of  the  power  of  which  it  is  the  cor- 
relative term.  Thus,  if  the  number  is  a  second  power,  the 
root  is  called  the  second  or  square  root;  if  it  is  the  third 
power,  the  root  is  called  the  third  or  cube  root ;  if  it  is  the 
fourth  power,  its  root  is  called  the  fourth  or  biquadrate  root ; 
and  so  on. 

Rational  roots  are  such  as  can  be  exactly  obtained. 

Surd  roots  are  such  as  cannot  be  exactly  obtained. 

519»  Roots  are  usually  denoted  by  writing  the  radical  sign, 
\/,  before  the  power,  with  the  index  of  the  root  over  it ;  in 
case,  however,  of  the  second  or  square  root,  the  index  2  is 
omitted.  Thus,  the  third  root  of  27  is  denoted  by  ^/27,  the 
second  root  of  16  is  denoted  by  \/16,  and  the  fourth  root  of  % 
is  denoted  by  /Z/%% 


372 


EVOLUTION. 


Roots  are  sometimes  denoted  by  a  fractional  index  or  expo- 
nent, of  which  the  numerator  indicates  the  power,  or  the  number 
of  times  the  number  is  to  be  taken  as  a  factor,  and  the  denomina- 
tor indicates  the  root,  or  the  number  of  equal  factors  into  which 
that  product  is  to  be  divided.  Thus  the  square  or  second  root 
of  12  is  denoted  by  12*,  the  fourth  root  of  f  by  (g)±,  and  the 
square  of  the  cube  root  of  27,  or  the  cube  root  of  the  square  of 
27,  is  denoted  by  275. 

520»  All  the  rational  roots  of  whole  numbers  are  also  whole 
numbers,  since  every  power  of  a  fractional  number  is  also  a 
fractional  number. 

521.  Prime  numbers  have  no  rational  roots. 

A  composite  number,  to  have  a  given  rational  root,  must 
have  the  exponent  of  the  power  of  each  of  its  prime  factors 
exactly  divisible  by  the  exponent  of  that  root. 

Note.  —  The  number  of  composite  numbers  that  have  rational  roots  is  com- 
paratively small.  The  number  of  rational  square  roots  of  whole  numbers 
from  1  to  250000  inclusive  is  only  500,  and  the  number  of  rational  cube  roots 
of  whole  numbers  from  1  to  8000000  inclusive  is  only  200. 

522 .  The  roots  represented  by  the  first  ten  numbers  and 
their  first  six  corresponding  powers  are  shown  in  the  following 

Table. 


1st  Power, 

1 

2 

3 

4 

5 

6 

7 

8 

9,     10 

2d  Power, 

1 

4 

9 

16 

25 

36 

49 

64 

81|    100 

3d  Power, 

1 

8 

27 

64 

125 

216 

343 

512 

729:   1000 

4th  Power, 

1 

1G 

81 

256 

625 

1296 

2401 

4096 

6561!  10000 

5th  Power, 

1 

82 

243 

1024 

3125 

7776 

16807 

32768 

59049'  100000 

6th  Power, 

1 

64 

729 

4096 

15625 

46656 

117649 

262144 

531441  1000000 

Note. — It  will  be  observed  by  the  table,  that  a  rational  square  root  can 
only  be  obtained  from  numbers  ending  in  1,  4,  5,  6,  or  9;  or  in  an  even 
number  of  ciphers,  preceded  by  one  of  these  figures.  It  is  true,  also,  that, 
if  the  square  number  ends  in  1,  its  square  root  ends  in  1  or  9 ;  if  in  4,  its  square 
root  ends  in  2  or  8 ;  if  in  9,  its  square  root  ends  in  3  or  7 ;  if  in  6,  its  square 
root  ends  in  4  or  6 ;  and  if  in  5,  its  square  root  ends  in  5. 

A  perfect  cube,  however,  may  end  in  either  of  the  nine  digits,  and  in  ciphers 
if  the  number  of  them  is  three  or  any  multiple  of  three;  also  if  the  cube  num- 
ber ends  in  1,  its  cube  root  will  end  in  1 ;  if  in  8,  its  cube  root  ends  in  2 ;  if  in 
3,  its  cube  root  ends  in  7 ;  if  in  4,  its  cube  root  ends  in  4 ;  if  in  5,  its  cube  root 
ends  in  5 ;  if  in  6,  its  cube  root  ends  in  6 ;  if  in  7,  its  cube  root  ends  in  3 ;  if  in 
8,  its  cube  root  ends  in  2 ;  and  if  in  9,  its  cube  root  ends  in  9, 


EVOLUTION.  373 

EXTRACTION  OF  THE  SQUARE  ROOT. 

523.  The  extraction  of  the  square  root  of  a  number  is  the 
process  of  finding  one  of  its  two  equal  factors  ;  or  of  finding 
such  a  factor  as,  when  multiplied  by  itself,  will  produce  the 
given  number. 

524  •  The  method  generally  adopted  for  extracting  the 
square  root  depends  upon  the  following  principles  :  — 

1.  The  square  of  any  number  has,  at  most,  only  twice  as 
many  figures  as  its  root,  and,  at  least,  only  one  less  than  twice 
as  many.  For  the  square  of  any  number  of  a  single  figure 
consists  of  either  one  or  two  places  of  figures,  as  l2  ==  1,  and 
92  =  81  ;  the  square  of  any  number  of  two  figures  consists  of 
either  three  or  four  places,  as  102  =  100,  and  992  ==  9801 ; 
and  the  same  law  holds  in  regard  to  numbers  of  three  or  more 
figures.  Therefore,  when  the  square  number  consists  of  one  or 
two  figures,  its  root  will  consist  of  one  figure ;  when  of  three  or 
four  figures,  its  root  will  consist  of  two  figures  ;  when  of  five  or 
six  figures,  its  root  will  consist  of  three  figures;  and  so  on. 
Hence,  if  a  number  be  separated  into  as  many  periods  as  pos- 
sible of  two  figures  each,  commencing  at  the  right,  to  these 
periods  respectively  will  correspond  the  units,  tens,  hundreds, 
&c.  of  the  square  root  of  the  number. 

2.  The  square  of  a  number  consisting  of  tens  and  units  is 
equal  to  the  square  of  the  tens,  plus  twice  the  product  of  the  tens 
into  the  units,  plus  the  square  of  the  units.  Thus,  if  the  tens  of 
a  number  be  denoted  by  a  and  the  units  by  b,  the  square  of  the 
number  will  be  denoted  by  (a  +  b)2  =  a2  +  2  a  b  +  b2. 
Then,  by  this  formula,  if  a  =  3,  and  5—6,  we  have  3  tens 
+  6  units  =  30  +  6  =  36;  and 

362  =  (30  +  6)2  =  302  +  2  X  (30  X  6)  X  62  =  1296. 
Or,  analytically, 


a  +  b    = 
a  +  b    = 

(a  +  b)  X  «  = 
(a  +  b)Xb  = 

30+6 
30+6 

302+30X6 

30X6+62 

=    30+     6          = 
=    SO-]-     6          h 

=  900+180          = 
=             180+36  = 

=  900+360+36  = 

36 
36 

1080 
216 

(a  +  b)2       m. 

302+2X(30X6)+62 

1296 

374  EVOLUTION. 

It  is  evident,  as  evolution  is  the  reverse  of  involution  (Art. 
517),  that  from  the  process  now  given  of  obtaining  a  square  may 
be  deduced  a  method  of  extracting  its  root.  Since  the  square  of 
(a-\-b)  is  a2  -|-  2  a  b  -\-  b2,  the  square  root  of  a2  -f-  2  a  b  -j-  b2 
must  be  a  -J-  b.  Now  it  will  be  observed  that  a,  the  first 
term  of  the  root,  is  the  square  root  of  a2,  the  first  term  of  the 
square  ;  and  if  a2  be  subtracted,  there  will  remain  2  a  b  -f-  b2, 
from  which  b,  the  second  term  of  the  root,  is  to  be  obtained. 
But  2  a  b  -\-  b2  is  the  same  as  (2  a  -\-  b)  \  b,  therefore  the 
remainder  equals  (2  a  -f-  b)  X  &•  But  as  b,  the  units,  is  al- 
ways much  less  than  2  a,  twice  the  tens,  we  consider  that  2  a 
X  b  is  about  equal  to  the  whole  remainder,  and  taking  2  a 
(which  we  know)  as  the  trial  divisor,  we  obtain  by  the  units. 
But  as  the  true  divisor  is  2  a  -\-  b,  we  add  the  units  to  twice 
the  tens  and  multiply  the  sum  by  the  units,  which  gives  a  pro- 
duct equal  to  the  whole  remainder,  or  2  a  b  -f-  b2. 

Since  every  number  of  more  than  one  figure  may  be  considered 
as  composed  of  tens  and  tin  its,  we  may  have  tens  and  units  of 
units,  tens  and  units  of  tens,  tens  and  units  of  hundreds,  &c. 
Hence,  the  principle  just  explained  applies  equally  whether  the 
root  contains  two  or  more  than  two  figures. 

525.    To  extract  the  square  or  second  root  of  numbers. 

Ex.  1.     What  is  the  square  root  of  1296  ?  Ans.  36. 

operation.  Beginning   at  the  right,   we  separate  the 

1  9  Q  c    qt  number  into  periods  of  two  figures  each,  by 

1  z  J  o    o  b  placing  a  point  (•)  over  the  right-hand  figure 

"  of  each  period.    Since  the  number  of  periods 

(56396  is   two,  the  root  will  consist  of  two  figures, 

3  o,  g  tens  and  units.     Then  1296  =  the  square  of 

the  tens  plus  twice  the  product  of  the  tens 


0  into  the  units,  plus  the  square  of  the  units 

The  square  of  tens  is  hundreds,  and  must 
therefore  be  found  in  the  hundreds  of  the  number.  The  greatest 
number  of  tens  whose  square  does  not  exceed  12  hundreds  is  3, 
which  we  write  as  the  tens  figure  of  the  root.  "We  subtract  the  9 
hundreds,  the  square  of  the  3  tens,  from  the  12  hundreds,  and  there 
remain  3  hundreds ;  after  which  we  write  the  figures  of  the  next 
period,  and  the  remainder  is  396  =  twice  the  product  of  the  tens 
into  the  units  plus  the  square  of  the  units.  We  have  then  next  to 
find  a  number  which,  added  to  twice  the  3  tens  of  the  root,  and  mul- 
tiplied into  their  sum,  shall  equal  396.     By  dividing  this  remainder 


EVOLUTION.  375 

by  twice  the  three  tens  of  the  root,  we  may  obtain  the  units,  a  num- 
ber somewhat  too  large.  But  though  it  may  be  too  large,  it  cannot  be 
too  small,  since  the  remainder  396  contains  twice  the  product  of  the 
tens  into  the  units,  and  also  the  square  of  the  units.  We  there- 
fore make  twice  the  three  tens  of  the  root  =  6  tens,  a  trial  divisor, 
with  which  we  divide  the  39  tens,  exclusive  of  the  6  units,  which 
cannot  form  any  part  of  the  product  of  the  tens  by  the  units.  The 
quotient  figure  obtained,  6,  must  be  the  units  figure  of  the  root,  or 
a  number  somewhat  larger.  To  determine  whether  it  expresses  the 
real  number  of  the  units  in  the  root,  we  annex  it  to  the  6  tens,  and 
multiply  the  number  66,  thus  formed,  by  it.  The  product  is  396, 
which  being  subtracted,  there  is  no  remainder.  Therefore  1296  is  a 
perfect  square,  and  36  the  root  sought. 

2.  What  is  the  square  root  of  278784?  Ans.  528. 

operation.  Since  there  are  three  pe- 


278784 
25 


102 

287 
204 

104* 

$ 

8384 
8384 

I  5  2  8  riods,  the  root  will   contain 

three  figures;  the  first  two 
may  be  considered  as  tens 
and  units  of  tens.  As  the 
square  of  tens  cannot  give 
less  than  hundreds,  we  must 
find  that  square  in  the  two 
left-hand  periods ;  and  as  we 
0  have  tens  and  units  of  tens, 

proof.  their  square  =  the  square  of 

528  X  528  ==  278784  tne  tens  Pms  tw^ce  tne  tens 

into  the  units,  plus  the  square 
of  the  units  =2787  (nearly).  We  proceed  then  with  the  first  two 
periods  exactly  the  same  as  when  the  root  consists  of  but  two  figures, 
and  thus  take  from  the  given  number  the  square  of  the  52  tens,  which 
leaves  a  remainder  of  8384.  We  now  consider  the  given  number 
278784,  as  the  square  of  a  number  consisting  of  52  tens  and  a  certain 
number  of  units,  which  square  will  of  course  equal  the  square  of  the 
tens  plus  twice  the  tens  into  the  units,  plus  the  square  of  the  units. 
But  the  square  of  the  tens,  or  (52)2  has  already  been  taken  from  the 
given  number,  leaving  a  remainder,  8384,  which  must  equal  twice  the 
tens  into  the  units  plus  the  square  of  the  units.  From  this  we  readily 
obtain  the  units,  just  as  when  we  had  but  two  figures  in  the  root. 

Rule.  —  Separate  the  given  number  into  as  many  periods  as  possible 
of  two  figures  each,  by  placing  a  point  over  the  place  of  units,  another 
over  the  place  of  hundreds,  and  so  on. 

Find  the  greatest  square  in  the  left-hand  period  ;  write  the  root  of  it 
at  the  right  of  the  given  number  after  the  manner  of  a  quotient  in  di- 
vision, and  subtract  the  second  power  from  the  left-hand  period. 

Bring  down  the  next  period  to  the  right  of  the  remainder  for  a  divi- 
dend, and  double  the  root  already  found  for  a  trial  divisor.  Find  hoio 
often  this  divisor  is  contained  in  the  dividend,  exclusive  of  the  right- 
hand  figure,  and  write  the  quotient  as  the  next  figure  of  the  root. 


376  EVOLUTION. 

Annex  the  last  root  figure  to  the  trial  divisor  for  the  true  divisor, 
which  multiply  by  the  last  root  figure  and  subtract  the  product  from  the 
dividend.  To  the  remainder  bring  down  the  next  period  for  a  new  div- 
idend. 

Double  the  root  already  found  for  a  new  trial  divisor,  and  continue 
the  operation  as  before,  till  all  the  periods  have  been  brought  down. 

Note  1.  —  When  the  product  of  any  trial  divisor  exceeds  its  corresponding 
dividend,  the  last  root  figure  must  be  made  less. 

If  a  dividend  does  not  contain  its  corresponding  divisor,  a  cipher  must 
be  placed  in  the  root,  and  also  at  the  right  of  the  divisor;  then,  after  bringing 
down  the  next  period,  this  last  divisor  must  be  used  as  the  divisor  of  the  new 
dividend. 

Note  2.  —  When  there  is  a  remainder  after  extracting  the  root  of  a  number, 
periods  of  ciphers  may  be  annexed,  and  the  figures  of  the  root  thus  obtained 
will  be  decimals. 

Note  3.  —  If  the  given  number  is  a  decimal,  or  a  whole  number  and  a  de- 
cimal, the  root  is  extracted  in  the  same  manner  as  in  whole  numbers,  except, 
in  pointing  off  the  decimals,  either  alone  or  in  connection  with  the  whole  num- 
ber, we  begin  at  the  separatrix  and  place  a  point  over  every  second  figure  to- 
ward the  right,  filling  the  last  period,  if  incomplete,  with  a  cipher.  The 
number  of  decimal  places  in  the  root  will  always  equal  the  number  of  periods 
of  decimals  in  the  power. 

Note  4.  —  If  the  given  number  is  a  common  fraction,  reduce  it  to  its  sim- 
plest form,  if  it  is  not  so  already,  and  extract  the  root  of  both  terms,  if  they  are 
perfect  powers ;  otherwise,  either  find  their  product,  extract  its  root,  and  di- 
vide the  result  by  the  denominator,  or  reduce  the  fraction  to  a  decimal,  and 
extract  the  root  of  the  decimal. 

Note  5.  —  When  the  given  number  is  a  mixed  number,  it  may  be  changed 
to  the  form  of  a  common  fraction,  or  the  fractional  part  may  be  reduced  to  a 
decimal,  before  attempting  to  extract  the  root. 

Examples. 
3.  What  is  the  square  root  of  T|-f^?  Ans.  ¥4T. 


4.  What  is  the  square  root  of  ¥3«y?  Ans.  .1936+. 

3  X  30  =  240;  ^2A0  =  .1936+. 

80  ' 


Or  &  =  .0375 ;  \/.0375  =  .1936+. 

5.  What  is  the  square  root  of  3444736  ?  Ans.  1856. 

6.  What  is  the  square  root  of  998001  ?  Ans.  999. 

7.  What  is  the  square  root  of  t3oVf  ? 

8.  Extract  the  square  root  of  234.09  ?  Ans.  15.3. 

9.  What  is  the  square  root  of  42£?  Ans.  6^. 


EVOLUTION. 


377 


10.  What  is  the  square  root  of  .000729  ? 

11.  What  is  the  square  root  of  17.3056? 

12.  What  is  the  square  root  of  52T9g  ? 

13.  What  is  the  square  root  of  95T*g-? 

14.  What  is  the  square  root  of  363^T  ? 

15.  How  much  is  \/1.96  ? 

16.  How  much  is  6561*  ? 

17.  How  much  is  \/93  ? 

18.  How  much  is  8*? 

19.  How    much    is 
9645192360241 ? 


one    of    the    two 


Ans.  .027. 
Ans.  4.16. 

Ans.  7  J. 

Ans.  9|. 
Ans.  19T1g-. 

Ans.  1.4. 

Ans.  81. 

Ans.  27. 

Ans.  64. 
equal    factors    of 
Ans.  3105671. 


526.  When  the  square  root  is  to  be  extracted  to  many 
places  of  figures,  the  work  may  be  contracted  thus  :  — 

Having  found  in  the  usual  ivay  one  more  than  half  of  the  root  figures 
required,  the  rest  may  be  found  by  dividing  the  last  remainder,  with  a 
single  figure  annexed  instead  of  two,  by  the  last  divisor,  proceeding  as 
in  contracted  division  of  decimals.     (Art.  276.) 


Examples. 

1.  What  is  the  square  root  of  785  to  five  places  of  decimals  ? 

Ans.  28.01785. 


CONTRACTED    METHOD. 


785 
4 


2  8.0  1  7  8  5+ 


48 


385 
384 


5601 


10000 
5601 


5602 


43990 
39219 


4771 

4482 

289 
280 

9 


COMMON    METHOD. 


785 
4 


2  8.0  1  7  8  5  -f 


48 

385 
384 

5601 

10000 
5601 

5602' 

7 

439900 
392189 

5602348 

4771100 
4482784 

560 

35 

6 

5 

28831600 
28017825 

813775 


The  nature  and  extent  of  the  contraction  will  be  seen  by  compar- 
ing the  contracted  method  with  the  common  method. 
32* 


378  EVOLUTION. 

2.  Extract  the  square  root  of  6§  to  four  places  of  deci- 
mals. Ans.  2.5298+. 

3.  Required  the  square  root  of  2  to  five  places  of  deci- 
mals. Ans.  1.41421+. 

4.  Required  the  square  root  of  3.15  to  eight  places  of  deci- 
mals. Ans.  1.77482393+. 

5.  Required  the  square  root  of  373  to  seven  places  of  deci- 
mals. Ans.  19.3132079+. 

6.  Extract  the  square  root  of  8.93  to  eight  places  of  deci- 
mals. Ans.  2.98831055+. 

EXTRACTION  OF  THE  CUBE  ROOT. 

527.  The  extraction  of  the  cube  root  of  a  number  is  the 
process  of  finding  one  of  its  three  equal  factors  ;  or,  of  finding 
a  factor  which,  being  multiplied  into  itself  twice,  will  produce 
the  given  number. 

528.  The  common  method  of  extracting  the  cube  root  de- 
pends upon  the  following  principles :  — 

1.  The  cube  of  any  number  has,  at  most,  only  three  times  as 
many  figures  as  its  root,  and,  at  least,  only  two  less  than  three 
times  as  many.  For  the  cube  of  a  number  of  a  single  figure 
consists  of,  at  most,  three  figures,  and,  at  least,  two  less  than 
that  number,  as  l3  =  1,  and  93  ess  729  ;  the  cube  of  a  number 
of  two  figures  consists  of,  at  most,  six  figures,  and,  at  least,  two 
figures  less  than  that  number,  as  103  =  1000,  and  993  = 
970299  ;  and  so  on.  Therefore,  when  a  cube  number  consists 
of  one,  two,  or  three  figures,  its  root  will  consist  of  one  figure ; 
when  of  four,  five,  or  six  figures,  its  root  will  consist  of  two 
figures,  and  so  on  ;  and  if  a  number  be  separated  into  as  many 
periods  as  possible  of  three  figures,  each  commencing  at  the 
right,  to  these  periods  respectively  will  correspond  the  units, 
tens,  hundreds,  &c.  of  the  cube  root  of  that  number. 

2.  Tlie  cube  of  a  number  consisting  of  tens  and  units  is 
equal  to  the  cube  of  the  tens,  plus  three  times  the  square  of  the 
tens  into  the  units,  plus  three  times  the  tens  into  the  square  of  the 
miits,  plus  the  cube  of  the  units.  'Jhus,  if  the  tens  of  a  number 
be  denoted  by  a,  and  the  units  by  b,  the  cube  of  the  number 


EVOLUTION.  379 

will  be  denoted  by  (a  +  b)3  ■=  a3  +  3  a2  b  +  3  a  b2  +  &3. 
Then,  by  this  formula,  if  a  =  3,  and  6  equal  6,  we  have  3  fens 
+  6  units  =  30  +  6  =  36,  and  3 63  ==  (30  +  6)3  =  303  +  3 
(302  X  6)  +  3  (30  X  62)  +  63  =  46656.     Or,  analytically, 

«  +  £       =30  +    6  =36 

a  _|-  6       =  30  +    6  =36 

(a-\-b)  Xa  =  302  +  30  X  6  =  To80 

(a+b)  Xb=  30  X  6-f62  =      216 

(a  +  6)2     =303  +  2  X30  X  6  +  62                       =    1296 
a_j_6      ,=  30  -j-6  .  = 36 

(a-Ji-b)2Xa  =  303  +  2  X  302  X  6  +  30  X  62  =  38880 

(a  +  6)2 X  b  = 3Q2  X  6  +  2X30X 68+68==  7776 

(a  +  b)3     =  303  +  3    (302X6)+3(30  X  62)+63=  46656 

It  is  evident,  as  evolution  is  the  reverse  of  involution,  that 
from  this  process  of  obtaining  a  cube  may  be  deduced  a 
method  of  extracting  the  cube  root.  Since  the  cube  of  a  +  b 
is  a3  +  3  a2  b  +  3  a  b2  +  b3,  the  cube  root  of  a3  +  3  a2  b  + 
3  a  b2  +  b3  must  be  a  +  J.  Now  a,  the  first  term  of  the  cube 
root,  is  the  cube  root  of  a3,  the  first  term  of  the  cube ;  and  if 
a3  be  subtracted,  there  will  remain  3  a2  b  +  3  a  b2  +  b3,  from 
which  b,  the  second  term  of  the  root,  is  to  be  obtained.  But 
3  a2  b  +  3  a  b2  +  b3  is  the  same  as  (3  a2  +  3  a  b  +  b2)  X  b  ; 
therefore  the  remainder  equals  (3  a2  +  3  a  b  +  b2)  X  &•  But 
as  3  a  b,  three  times  the  tens  into  the  units,  plus  b2,  the  square 
of  the  units,  is  generally  much  less  than  3  a2,  three  times  the 
square  of  the  tens,  we  consider  that  3  a2  X  b  is  about  equal  to 
the  whole  remainder,  and  taking  3  a2  (which  we  know)  as  the 
trial  divisor,  we  obtain  b,  the  units.  But  as  the  true  divisor  is 
3a2+  S  ab  -\~  b2  we  add  three  times  the  tens  by  the  units 
plus  the  square  of  the  units,  and  multiply  the  sum  by  the 
units,  which  gives  a  product  equal  the  whole  remainder,  or 
3aaft  +  3ay+  b3. 

Since  every  number  of  more  than  one  figure  may  be  con- 
sidered as  composed  of  tens  and  units,  we  may  have  tens  and 
units  of  units,  tens  and  units  of  tens,  tens  and  units  of  hundreds, 
&C.  Hence,  the  principle  just  explained  applies  equally  wheth- 
er the  root  contains  two  or  more  than  two  figures. 


380  EVOLUTION. 

529t     To  extract  the  cube  or  third  root  of  numbers. 
Ex.  1.     What  is  the  cube  root  of  46656  ?  Ans.  36. 

operation.  ^  Beginning    at 

46656 !  36  the  right,  we  sep- 

33  =  27  arate    the    given 


Trial  div.,  3  X  302  =  2700 

3  x  30  X  6  =    540 

62=      36 


True  divisor,  3276  X  6 


number  into  pe- 
196o6         riods,  by  placing 

a  point  over  the 

units  figure  and 
19656         over     tne     third 

figure  to  the  left. 


0         Since  the  number 
PR0°F*  ■  of  periods  is  two, 

36  X  36  X  36  =  46656  the  root  will  con- 

sist of  two  figures, 
tens  and  units.  Then  46656  =  the  cube  of  tens,  plus  three  times  the 
square  of  the  tens  into  the  units,  plus  three  times  the  tens  into  the 
square  of  the  units,  plus  the  cube  of  the  units.  The  cube  of  tens  is 
thousands,  and  must  therefore  be  found  in  the  thousands  of  the  num- 
ber. The  greatest  number  of  tens  whose  cube  does  not  exceed  46 
thousands  is  3,  which  we  write  as  the  tens  figure  of  the  root.  We 
then  subtract  the  27  thousands,  the  cube  of  the  3  tens,  from  the  46 
thousands,  and  there  remain  19  thousands;  and,  annexing  the  next 
period,  Ave  have  as  the  entire  remainder,  19656,  equal  three  times  the 
square  of  the  tens  into  the  units,  plus  three  times  the  tens  into  the 
square  of  the  units,  plus  the  cube  of  the  units,  or  the  product  of  three 
times  the  square  of  the  tens,  plus  three  times  the  tens  into  the  units, 
plus  the  square  of  the  units,  multiplied  by  the  units.  By  dividing 
this  remainder  by  three  times  the  square  of  the  tens  of  the  root,  we 
obtain  the  units,  or  a  number  somewhat  too  large.  Although  it  may 
be  too  large,  it  cannot  be  too  small,  since  the  remainder  19656  con- 
tains not  only  three  times  the  square  of  the  tens  into  the  units,  but 
three  times  the  tens  into  the  square  of  the  units,  plus  the  cube  of  the 
units.  We  therefore  make  three  times  the  square  of  the  tens  of  the 
root,  =  27  hundreds,  a  trial  divisor,  with  which  we  divide  the  196 
hundreds  of  the  remainder,  disregarding  the  56  units,  since  they 
cannot  form  any  part  of  the  product  of  the  square  of  the  tens  by  the 
units.  The  quotient  figure  obtained,  7,  must  be  the  units  figure  of 
the  root,  or  a  number  somewhat  larger. 

But  on  undertaking  to  complete  the  divisor  on  the  supposition  that 
7  is  the  true  units  figure  of  the  root,  we  find  a  divisor  too  large  for 
the  remainder.  We  therefore  take  6,  a  number  one  less,  and  to  de- 
termine whether  it  expresses  the  real  number  of  units  in  the  root,  we 
add  to  the  2  7  hundreds  of  the  trial  divisor  three  times  the  square  of 
the  3  tens  of  the  root  into  the  6  units,  plus  the  square  of  the  6  units ; 
and  multiplying  the  true  divisor,  3276,  thus  formed  by  the  units,  and 
subtracting  the  product,  19656,  from  the  remainder,  there  is  nothing 
left.     Hence,  46656  is  a  perfect  cube,  and  36  its  cube  root. 


EVOLUTION. 


381 


12326391 
8 


4326 


2.  What  is  the  cube  root  of  12326391  ? 

FIRST    OPERATION. 

23  = 
Trial  divisor,  3  X  202  =  1200 
3  X  20  X  3   ==    180 
32  =        9 

True  divisor,                        1389  X  3  =  JL 1 
Trial  divisor,  3  X  2302  &  158700 
3  X  230  X  1  =        G90 
12  = 1 

True  divisor,  159391  Xl  = 


Ans.  231. 

231 


15  9  3  9  1 


159391 


Since  there  are  three  periods  the  root  will  contain  three  figures, 
the  first  two  of  which  may  be  considered  as  tens  and  units  of  tens. 
As  the  cube  of  tens  cannot  give  less  than  thousands,  we  must  find 
that  cube  in  the  two  left-hand  periods ;  and  as  we  have  tens  and  units 
of  tens,  their  cube  will  equal  the  cube  of  the  tens,  plus  three  times  the 
square  of  the  tens  into  the  units,  plus  three  times  the  tens  into  the 
square  of  the  units,  plus  the  cube  of  the  units.  This  we  apply  to  the 
first  two  periods  exactly  the  same  as  when  the  root  consists  of  but  two 
figures;  and  thus  take  from  the  given  number  the  cube  of  the  23  tens, 
which  leaves  a  remainder  of  159391.  We  now  consider  the  given 
number,  159391,  as  the  cube  of  a  number  consisting  of  23  tens  and  a 
certain  number  of  units,  which  cube  will  of  course  equal  the  cube  of 
the  tens,  plus  three  times  the  square  of  the  tens  into  the  units,  plus 
three  times  the  tens  into  the  square  of  the  units,  plus  the  cube  of  the 
units.  But  the  cube  of  the  tens,  or  (23)3,  has  already  been  taken  from 
the  given  number,  leaving  a  remainder,  159391,  which  must  equal  three 
times  the  square  of  the  tens  into  the  units,  plus  three  times  the  tens 
into  the  square  of  the  units,  plus  the  cube  of  the  units  ;  or  equal  the 
product  of  three  times  the  square  of  23,  plus  three  times  23  into  the 
units,  plus  the  square  of  the  units,  multiplied  by  the  units.  From  this 
we  readily  obtain  the  units,  just  as  when  we  had  but  two  figures  in 
the  required  root. 


231 


SECOND    OPERATION. 

23 

— 

Trial  divisor, 
63X3 

=  1200 
=     189 

True  divisor, 
32 

sfcs  1389  x  3  = 
=          9 

Trial  divisor, 
691  X  1 

=  158700 

=t=            691 

True  divisor, 

1593  91  x  3r 

1-8 

8 

326291 

4326 

4167 

15  9  3  9  1 

159391 

382  EVOLUTION. 

In  the  second  operation  the  work  is  somewhat  abridged  by  render- 
ing shorter  the  method  of  finding  each  divisor  after  the  first,  of  both 
kinds.  Thus,  the  second  true  divisor  is  obtained  by  prefixing  to  the 
second  root  figure,  3,  three  times  the  part  of  the  root  preceding  it,  or 
3  X  2  ==  6,  and  adding  189,  the  product  of  the  number  63  thus 
formed  by  the  last  root  figure,  3,  to  the  preceding  trial  divisor,  or  1200 
-j-  189  =  1389,  second  true  divisor.  This  is  equivalent  to  adding 
*to  the  trial  divisor,  in  forming  the  true  divisor,  three  times  the  tens 
into  the  units,  plus  the  square  of  the  units,  or  3  X  20  X  3  +  32. 
For  the  second  trial  divisor  we  annex  two  ciphers  to  the  sum  found 
by  adding  together  the  square  of  the  last  root  figure,  the  last  true  di- 
visor, and  the  number  standing  over  it,  or  9  -f-  1389  -f-  189  =  1587, 
with  two  ciphers  annexed  =  158700,  the  second  trial  divisor.  This 
is  equivalent  to  taking  for  the  trial  divisor  three  times  the  square  of 
the  tens,  or  the  part  of  the  root  already  found,  or  3  X  230*.  The 
next  true  divisor  is  found  in  like  manner  as  was  the  second  true  di- 
visor. 

Rule  1.  —  Separate  the  given  number  into  as  many  periods  as  pos- 
sible of  three  figures  each,  b< ginning  at  the  units'  place. 

Find  the  greatest  cube  in  the  left-hand  period,  and  write  its  root  as 
the  first  figure  of  the  required  root.  From  that  period  subtract  the  cube, 
and  to  the  remainder  bring  down  the  next  period  for  a  dividend. 

Multiply  the  square  of  the  root  figure  by  3,  and  to  the  product  annex 
two  ciphers  for  a  trial  divisor,  and  see  how  often  it  is  contained  in  the 
dividend,  and  ivrite  the  result  as  the  next  figure  of  the  root. 

Add  to  the  trial  divisor  three  ti?nes  the  product  of  the  tens  figure  of  the 
root  by  the  units  figure  with  a  cipher  annexed,  and  the  square  of  the  last 
figureyfor  a  true  divisor. 

Multiply  the  true  divisor  by  the  last  figure  of  the  root ;  subtract  the 
product  from  the  dividend,  and  to  the  remainder  bring  down  the  next 
period  for  a  new  dividend. 

Multiply  the  square  of  the  root  figures  already  found  by  3,  and  to  the 
product  annex  two  ciphers  for  a  new  trial  divisor  ;  and  proceed  as  be- 
fore until  all  the  periods  are  brought  down.     Or, 

Rule  2.  — Having  found  the  first  trial  divisor  and  determined  the 
second  root  figure  as  by  the  preceding  rule, — 

Tale  three  times  the  part  of  the  root  already  found,  except  the  last 
figure,  to  it  annex  the  last  figure  of  the  root,  multiply  the  result  by  the 
figure  annexed,  and  write  the  product  below  the  trial  divisor,  and  add 
it  to  the  same  for  a  true  divisor. 

Multiply  the  true  divisor  by  the  last  figure  of  the  root :  subtract  the 
product  from  the  dividend,  and  to  the  remainder  bring  down  the  next 
period  for  a  new  dividend. 

To  the  last  true  divisor  and  the  number  immediately  over  it,  add  the 
square  of  the  last  root  figure,  and  to  the  sum  annex  two  ciphers  for  a 
new  trial  divisor ;  then  proceed  as  before. 

Note  1.  —  The  observations  made  in  Notes  1,  2,  3,  and  5,  under  the  rule  for 
the  extraction  of  the  square  root  (Art.  f>25),  are  equally  applicable  to  the  ex- 


EVOLUTION.  383 

traction  of  the  cube  root,  except  that  two  ciphers  must  be  placed  at  the  right 
of  a  true  divisor  when  it  is  not  contained  in  its  corresponding  dividend,  and  in 
pointing  off  decimals  each  period  must  contain  three  figures. 

Note  2.  —  If  the  given  number  is  a  common  fraction,  reduce  it  to  its  sim- 
plest form,  if  it  is  not  so  already,  and  extract  the  root  of  both  terms,  if  they  be 
perfect  powers;  otherwise,  either  find  the  product  of  the  numerator  by  the 
square  of  the  denominator,  extract  its  root,  and  divide  the  result  by  the  de- 
nominator; or  reduce  the  fraction  to  a  decimal,  and  extract  the  root  of  the 
decimal. 

Examples. 

3.  What  is  the  cube  root  of  77308776  ?  Ans.  426. 

4.  What  is  the  cube  root  of  Jf§?  Ans.  f. 

5.  What  is  the  cube  root  of  84.604519  ?  Ans.  4.39. 

6.  What  is  the  cube  root  of  54439939  ?  Ans.  379. 

7.  What  is  the  cube  root  of  60236288  ?  Ans.  392. 

8.  Required  the  cube  root  of  .726572699.  Ans.  .899. 

9.  Extract  the  third  root  of  109215352.  Ans.  478. 

10.  What  is  the  third  root  of  £§ $$  ?  Ans.  J-f . 

11.  What  is  the  value  of  ^ffjjif  ?  Ans.  f  f. 

12.  What  is  the  value  expressed  by  ^34965783  ? 

Ans.  327. 

13.  What  is  the  value  of  122615327232"  ?        Ans.  4968. 

14.  What  is  the  value  of  436036824287"  ?        Ans.  7583. 

530  •  When  the  cube  root  is  to  be  extracted  to  many  places 
of  decimals,  the  work  may  be  contracted  thus :  — 

Having  found  in  the  usual  way  one  more  than  half  of  the  root  figures 
required,  the  rest  may  he  found  by  dividing  the  last  remainder  by  its 
corresponding  true  divisor,  as  in  contracted  division  of  decimals  (Art. 
276),  observing  however  at  each  step  to  reject  two  figures  from  the  right 
of  the  divisor  and  one  from  the  right  of  the  remainder. 

Example. 

1.  Required  the  cube  root  of  2  to  four  places  of  decimals. 

Ans.  1.2599-f. 

2.  Find  the  third  root  of  11  to  four  places  of  decimals. 

Ans.  2.2239. 

3.  Extract  the  cube  root  of  3  to  six  places  of  decimals. 

Ans.  1.442249+. 

4.  Extract  the  cube  root  of  9  to  fifteen  places  of  decimals. 

Ans.  2.08008382301904. 


884 


EVOLUTION. 


EXTRACTION  OF  ANY  ROOT. 

531.  Tlie  root  corresponding  to  any  perfect  power  may  be 
obtained  by  resolving  that  power  into  its  prime  factors,  and  mul- 
tiplying together  one  of  each  number  of  equal  factors  denoted  by 
the  exponent  of  the  required  root.  Thus  one  of  each  two  equal 
factors  of  the  power  will  give  the  second  or  square  root ;  one 
of  each  three  equal  factors  wrill  give  the  third  or  cube  root ;  one 
of  each  four  equal  factors  will  give  the  fourth  root ;  and  so  on. 

532.  When  the  index  or  exponent  of  the  root  to  be  extracted 
is  a  composite  number,  the  root  may  be  obtained  by  successive 
extractions  of  the  simpler  roots  denoted  by  the  several  factors  of 
that  exponent.  Thus  the  fourth  root  may  be  obtained  by  ex- 
tracting the  square  root  twice  in  succession ;  the  sixth  root  by 
extracting  the  square  root  and  then  the  cube  root ;  and  so  on. 


K\AMI>LES. 

1.  Required  the  fourth  root  of  50625  ? 


Ans.  15. 


BY    FACTORS. 


BY    SUCCESSIVE    EXTRACTIONS. 


50625 


10125 


50025,225 
4 


52025 
X  5~|405 


4210G 
84 


445 


2225 

2225 


0 


381 

327 

3j9 

X~3 

5  X  3  =  15,  Ans. 

2.  What  is  the  square  root  of  998001  ? 

3.  What  is  the  cube  root  of  262144  ? 

4.  What  is  the  fourth  root  of  43046721  ? 

5.  What  is  the  fifth  root  of  14348907  ? 

6.  What  is  the  sixth  root  of  11390625  ? 


225 
1 

5125 
125 


15,  Ans. 


0 


Ans.  999. 
Ans.  64. 

Ans.  27. 
Ans.  15. 


533*  When  the  given  number  is  an  imperfect  power,  or 
otherwise,  or  the  exponent  denoting  the  root  is  prime,  or  other- 
wise, the  required  root  may  be  found  by  an  elegant  process, 
perfect  in  principle,  called  from  its  inventor, 


OPERATION. 

0 

0 

4 

16 

4 

16 

4 

32 

8 

4800 

4 

625 

120 

5425 

5 

650 

125 

607500 

5 

4059 

130 

611559 

5 

EVOLUTION.  385 

Horner's  Method. 

Ex.  1.  Kequired  the  cube  root  of  9295677.         Ans.  453. 

The  greatest  cube 
92959677  453  contained  in  the  left- 
£4  hand  period  we  find  to 

be    64,   whose   root,    4, 

28959  we  write    as    the    first 

27125  figure   of  the   required 

—— — ■  root.      This    figure,    4, 

±00^0/ /  we  wnte  under  the  ci- 

1834677  pher  of  the  first  column ; 

a  and    adding    it    to   the 

cipher  obtain  4,  which 
sum  multiplied  by  the  4  gives  16;  and 
the  result,  16,  we  write  under  the  cipher 
of  the  second  column,  and  by  addition 
obtain  16,  which  sum,  multiplied  by  the  4, 
gives  64  ;  and  the  result,  64,  we  write  in 
the  last  column  under  the  left-hand  period, 
92,  of  the  given  number.  The  64  sub- 
1350  tracted  from  the  92  above  it  gives  for  a  remainder  28. 

3  We   next   add  the  4  to  the  last  term,  4,  of  the  first 

0_0  column,  obtaining  8 ;  and  the  result,  8,  multiplied  by  the 

looo  4j  gives  32,  which  we  write  under  the  last  term,  16,  of 

the  second  column,  and,  adding  the  same  together,  obtain 
48.  We  next  add  the  4  to  the  last  term,  8,  of  the  first  column,  ob- 
taining 12  ;  and,  annexing  one  cipher  to  the  last  term,  12,  of  the  first 
column,  obtaining  120,  two  ciphers  to  the  last  term  of  the  second 
column,  obtaining  4800,  and  to  the  remainder  in  the  last  column 
bringing  down  the  next  period,  959,  obtaining  28959,  we  complete 
the  work  preparatory  to  the  finding  of  the  second  root  figure. 

To  determine  that  root  figure,  we  take  the  last  term,  4800,  of  the 
second  column,  for  a  trial  divisor,  and  the  last  term,  28959,  of  the 
last  column,  for  a  dividend;  and,  dividing,  6  would  appear  to  be  the 
second  figure  of  the  root.  This,  on  trial,  however,  is  found  to  be  too 
large ;  we  therefore  take  5,  which  answers.  This  5  we  add  to  the  last 
term,  120,  of  the  first  column,  obtaining  125  ;  which  sum,  125,  multi- 
plied by  the  5,  gives  625,  and  that  product,  added  to  the  last  term, 
4800,  of  the  second  column,  gives  5425  ;  and  this  result,  5425,  multi- 
plied by  the  5,  gives  27125,  which,  written  in  the  last  column  and  sub- 
tracted from  the  figures  above  it,  gives  a  remainder  1834.  Then 
we  add  the  5  to  the  last  term,  125,  of  the  first  column,  obtaining  130 ; 
and  the  result,  130,  multiplied  by  the  5,  gives  650,  which  we  write 
under  the  last  term,  5425,  of  the  second  column,  and  by  addition  ob- 
tain 6075.  We  next  add  the  5  to  the  last  term,  130,  of  the  first 
column,  obtaining  135  ;  and,  annexing  one  cipher  to  the  last  term, 
135,  of  the  first  column,  obtaining  1350,  two  ciphers  to  the  last  term, 
6075,  of  the  second  column,  obtaining  607500,  and  to  the  remainder 
N      33 


386  EVOLUTION. 

in  the  last  column  bringing  down  the  next  period,  677,  obtaining 
1834677,  the  work  is  completed  preparatory  to  finding  the  third  root 
figure. 

To  determine  that  figure,  as  before,  we  divide  the  last  term  of  the 
third  column  by  the  last  term  of  the  second.  We  thus  obtain  3, 
which,  added  to  the  last  term  of  the  first  column,  gives  1353,  which 
sum,  multiplied  by  the  3,  gives  4059  ;  and  that  product,  being  added 
to  the  last  term  of  the  second  column,  gives  611559  ;  and  that  sum, 
multiplied  by  the  3,  gives  1834677,  which  being  exactly  as  large  as 
the  last  term  of  the  third  column,  on  being  written  under  it  and  sub- 
tracted there  is  no  remainder.  The  given  number  is  therefore  a  per- 
fect power,  and  the  cube  root  sought  is  453. 

In  practice,  the  work  may  be  performed  with  less  figures,  by  writ- 
ing down  in  the  several  columns  only  the  results. 

Rule. —  Commence  as  many  columns  as  there  are  units  in  the  ex- 
ponent of  the  root  to  be  extracted,  by  writing  the  given  number  as  the 
head  of  the  right-hand  column,  and  a  cipher  as  the  head  of  each  of  the 
others. 

Separate  the  given  number  into  as  many  periods  as  possible  of  as 
many  figures  each  as  the  exponent  of  the  root  requires ;  and  having 
found  the  nearest  root  of  the  left-hand  period,  write  it  as  the  first  figure 
of  the  required  root. 

Write  this  figure  in  the  first  column,  and,  having  added  it  to  what 
stands  above  it,  multiply  the  sum  by  the  same  figure,  and  write  the  pro- 
duct in  the  second  column  ;  add,  in  like  manner,  in  the  second  column, 
and  multiply  the  sum  by  the  same  figure,  writing  the  product  in  the  third 
column;  and  so  proceed,  writing  the  last  product  in  the  last  column,  and 
subtracting  it  from  what  stands  above  it. 

Then  add  the  same  figure  to  the  last  term  of  the  first  column,  multi- 
ply the  sum  by  the  same  figure,  and  add  the  product  to  the  last  term  of 
the  second  column;  and  so  on,  writing  the  last  product  in  the  last 
column  but  one.  Repeat  the  process,  stopping  each  time  with  one  col- 
umn farther  to  the  left,  till  the  last  product  shall  fall  in  the  second 
column-. 

Add  the  figure  found  for  the  root  to  the  last  term  of  the  first  column ; 
annex  one  cipher  to  the  last  number  in  the  first  column,  two  ciphers  to 
the  last  number  in  the  second  column,  and  so  on ;  and  to  the  last  num- 
ber in  the  last  column  bring  down  the  next  period  for  a  dividend. 

Take  the  last  term  of  the  column  next  to  the  last  for  a  trial  divisor, 
and  see  how  often  it  is  contained  in  the  dividend,  and  write  the  result  as 
the  next  figure  of  the  root. 

Add  'this  figure  to  the  last  term  of  the  first  column,  multiplying  the 
sum  by  the  same  figure,  add  the  product  to  the  second  column,  and 
so  on ;  proceed  as  before,  till  all  the  periods  have  been  brought  down,  or 
an  answer  sufficiently  exact  has  been  obtained. 

Note  1.  —  When  any  dividend  will  not  contain  its  corresponding  trial  divi- 
sor, write  a  cipher  in  the  root,  bring  down  to  the  dividend  another  period,  an- 
nex an  additional  cipher  to  the  last  term  of  the  first  column,  two  additional 


EVOLUTION. 


387 


ciphers  to  the  last  term  of  the  second  column,  and  so  on;  and  use  the  same 
trial  divisor  as  before,  increased  however  by  the  additional  ciphers. 

Note  2.  —  When  the  given  number  does  not  have  an  exact  root,  periods  of 
ciphers  may  be  annexed. 

Note  3.  —  When  the  root  is  required  to  many  places  of  decimals,  the  work 
may  be  contracted  by  rejecting  one  figure  at  the  right  from  the  number  in  the 
column  next  to  the  last,  two  from  the  number  in  the  column  next  farther  to 
the  left,  and  so  on,  and  otherwise  proceeding  as  directed  in  the  rule,  except  that 
the  new  figure  of  the  root  is  not  added  to  the  first  column.  As  soon  as  all  the 
figures  are  rejected  in  the  number  of  the  first  column,  the  remainder  of  the 
work  may  be  performed  as  in  contracted  division  of  decimals  (Art.  276). 

Examples. 

2.  Required  the  fourth  root  of  1.016397  to  eight  places  of 
decimals.  Ans.  1.00407427. 


OPERATION. 

0 

1 

0 
1 

1 
2 

3 
3 

0 

1 
1 

3 

1.0163970011.00407427 
1 

1 
1 

1639700 
1609632 

2 
1 

4000000 
2408- 

30068 
28338 

3 
1 

60000 
2- 

402408 
2416 

1730 
1619 

400 

602 

2 

"604 

2 

606 

404824 

111 
81 

30 

28 

~2 

3.  What  is  the  cube  root  of  41673648563  ?       Ans.  3467. 

4.  What  is  the  cube  root  of  43614208? 

5.  What   is  the  cube  root  of  1.05  to  six  places  of  deci- 
mals? Ans.  1.016397. 

6.  What  is  the  fifth  root  of  184528125  ?  Ans.  45. 

7.  Required  the  fourth  root  of  100  to  six  places  of  deci- 
mals ?  Ans.  3.162278. 

8.  Required  the  fifth  root  of  the  fourth  power  of  9  to  seven 
places  of  decimals?  Ans.  5.7995466. 


388 


EVOLUTION. 


APPLICATIONS    OF  POWERS  AND   ROOTS. 

534.  A  Triangle  is  a  figure  having  three  sides  and  three 
angles.  When  one  of  the  sides  of  a  triangle  is  perpendicular 
to  another  side,  the  opening  between  them  is  called  a  right 
angle,  and  the  triangle  is  called  a  right-angled  triangle. 

The  lowest  side,  A  B,  is  called  the  base  of 
the  triangle  ABC,  the  side  B  C  the  perpen- 
dicular, the  longest  side,  A  C,  the  hypothenuse, 
and  the  angle  at  B  is  a  right  angle.  Also,  the 
line  B  C,  being  perpendicular  to  the  base,  is 
the  altitude.  Base. 

535.  The  square  described  upon  the  hypothenuse  of  a  right 
angled  triangle  is  equivalent  to  the  sum  of  the  squares  de- 
scribed upon  the  other  two  sides. 


Tims,  if  the  hypothenuse  A  C  be  5  feet, 
the  base  AB  4  feet,  and  the  perpendicular 
B  C  3  feet,  then  5a  =  4*  -f-  3'2,  or  25  = 
16  -J-  9. 


B 

536.     To  find  the  hypothenuse,  the  base  and  perpendicular 
being  given. 

Add  together  the  square  of  the  base  and  the  square  of  the  perpendicu- 
lar, and  extract  the  square  root  of  the  sum. 

Thus,  if  the  base  be  4  and  the  perpendicular  3,  the  hypothenuse 
will  equal  V4*  -J-  32  =  \/lb  =  5. 

537 #     To  find  the  perpendicular,  the  base  and  hypothenuse 
being  given. 

Subtract  the  square  of  the  base  from  the  square  of  the  hypothenuse, 
and  extract  the  square  root  of  the  remainder. 

Thus,  if  the  base  be  4  and  the  hypothenuse  5,  the  perpen- 
dicular will  equal  */  5*  —  42  =s  /^/l)-—  3. 

538.     To  find  the  base,  the  hypothenuse  and  perpendicular 
being  given. 


EVOLUTION. 


389 


; 


Subtract  the  square  of  the  perpendicular  from  the  square  of  the  hy- 
pothenuse,  arid  extract  the  square  root  of  the  remainder. 

Thus,  if  the  perpendicular  be  3  and  the  hypothenuse  5,  the 
base  will  equal  „/  5"  — &  =  VT6  —  4. 

539  •     All  triangles  having  the  same  base  are  to  each  other 
as  their  altitudes. 

All  similar  triangles,  and  other  similar  rectilineal  figures,  are 
to  each  other  as  the  squares  of  their  homolo- 
gous or  corresponding  sides. 

Thus,  the  triangles  ACE  and  A  C  D, 
having  the  same  base,  A  C,  are  to  each  other 
as  the  altitude  E  C  of  the  one  is  to  the  alti- 
tude D  C  of  the  other. 

Also,  the  triangles  ACE  and  BCD, 
having  their  corresponding  angles  the  same,  and  their  sides  in 
direct  proportion,  are  said  to  be  similar,  and  are  to  each  other 
as  the  squares  of  their  corresponding  sides,  or  as  (A  E)2  is  to 
(B  D)2,  (A  C)2  is  to  (B  C)2,  and  (C  E)2  is  to  (C  D)2.  Like- 
wise the  larger  square,  of  which  A  C  is  one  of  the  equal  sides, 
is  to  the  smaller  square,  of  which  B  C  is  one  of  the  equal  sides, 
as  (A  C)2  is  to  (B  C)2. 

540.  All  circles  (Art.  143)  are  to  each  other  as  the  squares 
of  their  diameters,  semidiameters,  or  circumferences. 

The  circumference  of  a  circle  is  the  line 
which  bounds  it ;  and  the  diameter  is  a  line 
drawn  through  the  centre,  and  terminated 
by  the  circumference ;  as  A  B  and  C  D. 

Then,  the  larger  circle,  of  which  A  B  is 
the  diameter,  is  to  the  smaller,  of  which  C  D 
is  the  diameter,  as  (A  B)2  is  to  (C  D)2,  &c. 

541.  To  find  the  side,  diameter,  or  circumference  of  any 
surface,  which  is  similar  to  a  given  surface. 

State  the  question  as  in  Proportion,  and  square  the  given  sides,  diam- 
eters, or  circumferences,  and  the  square  root  of  the  fourth  term  of  the 
proportion  will  be  the  answer  required. 

Thus,  if  12  feet  be  the  length  of  a  side  of  a  triangle  whose 
area  is  72  square  feet,  the  length  of  the  corresponding  side  of  a 
33* 


890  EVOLUTION. 

similar  triangle  whose  area  is  32  square  feet  would  be  found 
as  follows : 
72  :  32  :  :  122  =*  144  :  64 ;  y^T  £s  8  feet,  length  required. 

542.  To  find  the  area  of  any  surface  which  is  similar  to  a 

given  surface. 

State  the  question  as  in  Proportion,  and  square  the  given  .sides,  diam- 
eters, or  circumferences,  and  the  fourth  term  of  (lie  proportion  will  be  the 
answer  required. 

Thus,  if  72  square  feet  be  the  area  of  a  triangle  of  which  12 
feet  is  one  of  the  sides,  the  area  of  a  similar  triangle  of  which 
the  corresponding  side  is  8  feet  would  be  found  as  follows : 

122  =  144  :  82  =  64  :  :  72  sq.  ft.  :  32  sq.  ft.,  area  required. 

543.  To  find  the  side  of  a  square  equal  in  area  to  any  given 

surface. 

Find  the  square  root  of  tlie  given  area,  and  that  root  ivill  be  the  side 
of  the  area  required. 

544.  A  sphere  is  a  solid  bounded  by  a  continued  convex 
surface,  every  part  of  which  is  equally  distant  from  the  point 
within  called  the  centre. 

A 

The  diameter  of  a  sphere  is  a  straight  line  pass- 
ing through  the  centre,  and  terminated  by  the  sur- 
face ;  as  A  B. 


o 


545.  A  cone  is  a  solid  having  a  circle  for  its  base,  and 
tapering  uniformly  to  a  point,  called  the  vertex. 

The  altitude  of  a  cone  is  its  perpendicular  height, 
or  a  line  drawn  from  the  vertex  perpendicular  to 
the  plane  of  the  base,  as  B  C.  The  diameter  of  its 
base  is  a  straight  line  drawn  through  the  centre  of 
the  plane  of  the  base  from  one  side  of  the  circle  to 
the  other ;  as  A  D. 

546.  Spheres  are  to  each  other  as  the  cubes  of  their  diam- 
eters, or  of  their  circumferences. 

Similar  cones  are  to  each  other  as  the  cubes  of  their  altitudes, 
or  of  the  diameters  of  their  bases. 

All  similar  solids  are  to  each  other  as  the  cubes  of  their 
homologous  or  corresponding  sides,  or  of  their  diameters. 


EYOLUTION.  391 

Note.  —  Cones  and  other  solids  are  said  to  be  similar  when  their  corre- 
sponding parts  are  in  direct  proportion  to  each  other. 

547.  To  find  the  contents  of  any  solid  which  is  similar  to  a 
given  solid. 

State  the  question  as  in  Proportion,  and  cube  the  given  sides,  diam- 
eters, altitudes,  or  circumferences,  and  the  fourth  term  of  the  -proportion 
is  the  answer  required. 

548.  To  find  the  side,  diameter,  altitude,  or  circumference 
of  any  solid,  which  is  similar  to  a  given  solid. 

State  the  question  as  in  Proportion,  and  cube  the  given  sides,  diam- 
eters, altitudes,  or  circumferences,  and  the  cube  root  of  the  fourth  term 
of  the  proportion  is  the  answer  required. 

549.  To  find  the  side  of  a  cube  that  shall  be  equal  in 
solidity  to  any  given  solid. 

Find  the  cube  root  of  the  contents  of  the  given  solid,  and  that  root  will 
be  the  side  of  the  cube  required. 

550.  To  find  a  mean  proportional  (Art.  333)  between  any 
two  numbers. 

Find  the  square  root  of  the  product  of  the  two  numbers,  and  that  root 
will  be  the  mean  proportional  required. 

551.  To  find  two  mean  proportionals  between  two  given 
numbers. 

Find  the  cube  root  of  the  quotient  of  the  greater  of  the  two  numbers 
divided  by  the  less.  The  product  of  the  less  number  by  that  root  icill  be 
the  least  mean  proportional ,  and,  the  quotient  of  the  greater  number  by 
the  same  root  will  be  the  other  mean  proportional. 

552.  To  find  any  two  numbers,  whose  sum  and  product  are 

given. 

From  the  square  of  half  the  sum  of  the  two  numbers  subtract  their 
product,  and  the  square  root  of  the  remainder  will  equal  half  the  differ- 
ence of  the  two  numbers,  which  added  to  half  their  sum  will  give  the 
larger,  and  subtracted  from  half  their  sum  will  give  the  smaller,  of  the 
numbers  required, 

558.  To  find  any  two  numbers,  when  their  sum  and  the 
difference  of  their  squares  are  given. 

The  difference  of  their  squares  divided  by  the  sum  of  the  numbers  will 
give  their  difference;  and  half  of  their  difference  added  to  half  of  their 
sum  will  give  the  larger,  and  half  of  their  difference  subtract 
half  of  their  sum  icill  give  the  smaller,  of  the  numbers  required. 


I 

392  EVOLUTION. 

Examples. 

1.  A  certain  general  has  an  army  of  141376  men.  How 
many  must  he  place  in  rank  and  file  to  form  them  into  a 
square?  Ans.  376. 

2.  If  the  area  of  a  circle  be  1760  yards,  how  many  feet  must 
the  side  of  a  square  measure  to  contain  that  quantity  ? 

Ans.  125.857+  feet 

3.  If  a  line  144  feet  long  will  reach  from  the  top  of  a  fort 
to  the  opposite  side  of  a  river  64  feet  wide,  on  whose  brink  it 
stands,  what  is  the  height  of  the  "fort  ?  Ans.  128.99-)-. 

4.  A  certain  room  is  20  feet  long,  16  feet  wide,  and  12  feet 
high  ;  how  long  must  a  line  be  to  extend  from  one  of  the  lower 
corners  to  the  upper  corner  farthest  from  it  ?      Ans.  28.28ft\ 

5.  A  certain  field  is  40  rods  square  ;  what  must  be  the  length 
of  one  of  the  equal  sides  of  another  field  that  shall  contain  only 
one  fourth  as  much  area  ?  Ans.  20  rods. 

6.  The  areas  of  two  similar  triangular-shaped  fields  are  60 
and  90  acres,  and  a  side  of  the  former  is  66  rods.  Required 
the  corresponding  side  of  the  latter  ? 

7.  If  a  lead  pipe  £  of  an  inch  in  diameter  will  fill  a  cistern 
in  3  hours,  what  should  be  its  diameter  to  fill  it  in  2  hours  ? 

Ans.  .918-f-  inches. 

8.  If  a  pipe  1J-  inches  in  diameter  will  fill  a  cistern  in  50 
minutes,  how  long  would  it  require  a  pipe  that  is  2  inches  in 
diameter  to  fill  the  same  cistern  ?  Ans.  28m.  7^s. 

9.  If  a  pipe  6  inches  in  diameter  will  draw  off  a  certain 
quantity  of  water  in  4  hours,  in  what  time  would  it  take  3  pipes 
of  four  inches  in  diameter  to  draw  off  twice  the  quantity  ? 

Ans.  6  hours. 

10.  The  first  term  of  a  proportion  is  40,  and  the  fourth  term 
90.    Required  a  mean  proportional  between  them.    Ans.  60. 

11.  In  a  pair  of  scales  a  body  weighed  31£  pounds  in  one 
scale,  and  only  20  pounds  in  the  other  scale.  Required  its  true 
weight.  Ans.  25  pounds. 

12.  I  wish  to  set  out  an  orchard  of  2400  mulberry-trees,  so 
that  the  length  shall  be  to  the  breadth  as  3  to  2,  and  the  dis- 
tance between  any  two  adjacent  trees  7  yards.  How  many 
trees  must  there   be  in  the  length,   and   how   many  in   the 


EVOLUTION.  393 

breadth ;  and  on  how  many  square  yards  of  ground  will  they 
stand  ?      Ans.   GO  in  length  ;  40  in  breadth  ;  112749  sq.  yd. 

13.  The  sum  of  two  persons' ages  is  50  years,  and  their  pro- 
duct is  600  years.     What  are  their  ages  ? 

Ans.  Of  the  one,  20  years ;  of  the  other,  30  years. 

14.  Two  ships  sail  from  the  same  port ;  one  goes  due  north 
128  miles,  the  other  due  east  72  miles;  how  far  are  the  ships 
from  each  other?  Ans.  146.86-]— 

15.  There  are  two  columns  in  the  ruins  of  Persepolis  left 
standing  upright ;  one  is  70  feet  above  the  plane,  and  the  other 
50 ;  in  a  straight  line  between  these  stands  a  small  statue,  5  feet 
in  height,  the  head  of  which  is  100  feet  from  the  summit  of  the 
higher,  and  80  feet  from  the  top  of  the  lower  column.  Re- 
quired the  distance  between  the  tops  of  the  two  columns. 

16.  The  sum  of  two  numbers  is  44,  and  the  square  of  their 
difference  is  16.     Required  the  numbers. 

Ans.  24  the  larger  number  ;  20  the  smaller. 

17.  A  tree  80  feet  in  height  stands  on  a  horizontal  plane ;  at 
what  height  from  the  ground  must  it  be  broken  off,  so  that  the 
top  of  it  may  fall  on  a  point  40  feet  from  the  bottom  of  the  tree, 
the  end  where  it  was  broken  off  resting  on  the  stump  ? 

Ans.  30  feet. 

18.  The  height  of  a  tree,  growing  in  the  centre  of  a  circular 
island,  100  feet  in  diameter,  is  160  feet;  and  a  line  extending 
from  the  top  of  it  to  the  farther  shore  is  400  feet.  What  is  the 
breadth  of  the  stream,  provided  the  land  on  each  side  of  the 
water  be  level?  Ans.  316.6-)- feet. 

19.  A  ladder  70  feet  long  is  so  planted  as  to  reach  a  win- 
dow 40  feet  from  the  ground,  on  one  side  of  the  street,  and 
without  moving  it  at  the  foot  it  will  reach  a  window  30  feet  high 
on  the  other  side  ;  what  is  the  breadth  of  the  street  ? 

20.  If  an  iron  wire  -j-V  of  an  inch  in  diameter  will  sustain  a 
weight  of  450  pounds,  what  weight  might  be  sustained  by  a 
wire  an  inch  in  diameter  ?  Ans.  450001b. 

21.  A  gentleman  proposes  to  plant  a  vineyard  of  10  acres. 
If  he  places  the  vines  6  feet  apart,  how  many  more  can  he 
plant  by  setting  them  in  the  quincunx  order  than  in  the  square 
order,  allowing  the  plat  to  lie  in  the  form  of  a  square,  and  no 
vine  to  be  set  nearer  its  edge  than  1  foot  in  either  case  ? 

Ans.  1870  more  in  the  quincunx  order. 


■ 
394  EVOLUTION. 

22.  Four  men,  A,  B,  C,  and  D,  bought  a  grindstone,  the  dia- 
meter of  which  was  40  inches  and  the  place  for  the  shaft  4 
inches  in  diameter.  It  was  agreed  that  A  should  grind  off  his 
share  first,  then  in  turn  B,  C,  and  D.  Required  how  many 
inches  each  man  will  grind  off  from  the  semidiameter,  provid- 
ing they  each  paid  the  same  sum. 

Ans.  A,  2.651in. ;  B,3.137in. ;  C,  4.064in. ;  and  D,  8.148in. 

23.  I  have  a  board  whose  surface  contains  49 §■  square  feet; 
the  board  is  1£  inches  thick,  and  I  wish  to  make  a  cubical  box 
of  it.     Required  the  length-  of  one  of  its  equal  sides. 

Ans.  36  inches. 

24.  A  carpenter  has  a  plank  1  foot  wide,  22^J  feet  long, 
and  2J-  inches  thick ;  and  he  wishes  to  make  a  box  whose 
width  shall  be  twice  its  height,  and  whose  length  shall  be  twice 
its  width.     Required  the  contents  of  the  box. 

Ans.  5719  cubic  inches. 

25.  If  a  ball,  3  inches  in  diameter,  weigh  4  pounds,  what 
will  be  the  weight  of  a  ball  that  is  G  inches  in  diameter  ? 

Ans.  321bs. 

26.  If  a  globe  of  gold,  one  inch  in  diameter,  be  worth  $  120, 
what  is  the  value  of  a  globe  3£  inches  in  diameter? 

27.  If  the  weight  of  a  well-proportioned  man,  5  feet  10 
inches  in  height,  be  180  pounds,  what  must  have  been  the 
weight  of  Goliath  of  Gath,  who  was  10  feet  4§  inches  in 
height?  Ans.  1015.1-f-lb. 

28.  If  a  bell,  4  inches  in  height,  3  inches  in  width,  and  \  of 
an  inch  in  thickness,  weigh  2  pounds,  what  should  be  the 
dimensions  of  a  similar  bell  that  would  weigh  2000  pounds  ? 

Ans.  3ft.  4in.  high,  2ft.  6in.  wide,  and  2^in.  thick. 

29.  What  are  the  two  mean  proportionals  between  56  and 
12096?  Ans.  336  and  2016. 

30.  Having  a  small  stack  of  hay,  5  feet  in  height,  weighing 
lcwt,  I  wish  to  know  the  weight  of  a  similar  stack  that  is  20 
feet  in  height.  Ans.  64cwt. 

31.  If  a  man  dig  a  small  square  cellar,  which  will  measure 
6  feet  each  way,  in  one  day,  how  long  would  it  take  him  to  dig 
a  similar  one  that  measured  10  feet  each  way  ? 

Ans.  4.629+  days. 

32.  If  an  ox,  whose  girth  is  6  feet,  weighs  6001b.,  what  is 
the  weight  of  an  ox  whose  girth  is  8  feet?   Ans.  1422.2-{-lb. 


PROGRESSION,   OR   SERIES.  395 

33.  Four  women  own  a  ball  of  yarn,  5  inches  in  diameter. 
It  is  agreed  that  each  shall  wind  off  her  share  from  the  ball. 
How  many  inches  of  its  diameter  shall  each  wind  off? 

Ans.  First,  .45-f-  inches  ;  second,  .57-f-  inches  ;  third,  .82— j— 
inches  ;  fourth,  3.149-(-  inches. 

34.  John  Jones  has  a  stack  of  hay  in  the  form  of  a  quad- 
rangular pyramid.  It  is  16  feet  in  height,  and  12  feet  wide  at 
its  base.  It  contains  5  tons  of  hay,  worth  $  17.50  per  ton. 
Mr.  Jones  has  sold  this  hay  to  Messrs.  Pierce,  Row,  Wells,  and 
Northend.  As  the  upper  part  of  the  stack  has  been  injured,  it 
is  agreed  that  Mr.  Pierce,  who  takes  the  upper  part,  shall  have 
10  per  cent,  more  of  the  hay  than  Mr.  Rowe  ;  and  Mr.  Rowe, 
who  takes  his  share  next,  shall  have  8  per  cent,  more  than 
Mr.  Wells  ;  and  Mr.  Northend,  who  has  the  bottom  of  the 
stack,  that  has  been  much  injured,  shall  have  10  per  cent, 
more  than  Mr.  Wells.  Required  the  quantity  of  hay,  and 
how  many  feet  of  the  height  of  the  stack,  beginning  at  the  top, 
each  receives. 

Ans.  Pierce  receives  27^/TjCwt. and  10.3 6 6-|- feet  in  height; 
Rowe,  24j-f|cwt.  and  2.493  feet;  Wells,  22§4Jcwt.  and 
1.666  feet;  Northend,  2o2\\cwt.  and  1.474  feet. 


PROGRESSION,  OR  SERIES. 

554.  A  Series  is  a  succession  of  numbers  that  depend  on 
one  another  by  some  fixed  law. 

The  numbers  constituting  a  series  are  called  its  terms;  of 
which  the  first  and  last  are  called  extremes,  and  the  other 
terms  the  means. 

ARITHMETICAL  PROGRESSION. 

555.  Arithmetical  Progression,  or  Progression  by 
Difference,  is  a  series  that  increases  or  decreases  by  a  con- 
stant number,  called  the  common  difference. 

The  series  is  said  to  be  an  ascending  one  when  each  terra 


396  PROGRESSION,  OR  SERIES. 

after  the  first  exceeds  the  one  before  it ;  and  a  descending  one 
when  each  term  after  the  first  is  less  than  the  one  before  it. 

Thus,  1,  5,  9,  13,  17,  21,  25,  29,  33,  is  an  ascending  series, 
in  which  each  term  after  the  first  is  derived  from  the  one  pre- 
ceding it  by  the  addition  of  the  common  difference  4 ;  and  25, 
22,  19,  16,  13,  10,  7,  4,  1,  is  a  descending  series,  in  which  each 
term  after  the  first  is  derived  from  the  one  preceding  it  by  the 
subtraction  of  the  common  difference  3. 

'556.  In  arithmetical  progression,  the  first  term,  the  last 
term,  the  number  of  terms,  the  common  difference,  and  the  sum 
of  the  terms,  are  so  related  to  each  other,  that,  three  of  these  be- 
ing given,  the  two  others  may  be  readily  determined. 

557  •  To  find  the  common  difference,  the  extremes  and  num- 
ber of  terms  being  given. 

Ex.  1.  The  extremes  of  an  arithmetical  series  are  3  and 
45,  and  the  number  of  terms  is  22.  Required  the  common  dif- 
ference. Ans.  2. 

operation.  It  is  evident  that  the  number  of  common 

45 g  differences  in  any  series  must  be  1  less  than 

. =  2.       the  number  of  terms.     Therefore,  since  the 

22  —  1  number  of  terms  in  this  series  is  22,  the  num- 

ber of  common  differences  will  be  22  —  1  = 
21,  and  their  sum  will  be  equal  to  the  difference  of  the  extremes; 
hence  the  difference  of  the  extremes,  45  —  3  =  42,  divided  by  the 
number  of  common  differences,  21,  gives  2  as  the  common  difference 
required. 

Rule. — Divide  the  difference  of  the  extremes  by  the  number  of 
terms  less  one,  and  the  quotient  ivill  be  the  common  difference. 

Examples. 

2.  A  certain  school  consists  of  19  teachers  and  scholars, 
whose  ages  form  an  arithmetical  series ;  *the  youngest  is  3 
years  old,*  and  the  oldest  39.  What  is  the  common  difference 
of  their  ages  ?  Ans.  2  years. 

3.  A  man  is  to  travel  from  Albany  toV  certain  place  in  11 
days,  and  to  go  but  5  miles  the  first  day,  increasing  the  distance 
equally  each  day,  so  that  the  last  day's  journey  may  be  45 
miles.     Required  the  daily  increase.  Ans.  4  miles. 


PROGRESSION,   OR   SEBIB8.  897 

558.  To  find  the  number  of  terms,  the  extremes  and  com- 
mon difference  being  given. 

Ex.  1.  If  the  extremes  of  an  arithmetical  series  are  3  and 
19,  and  the  common  difference  is  2,  what  is  the  number  of 
terms  ?  Ans.  9. 

operation.  It  is  evident,  that,  if  the  difference  of  the 

19  —  3  extremes  be  divided  by  the  common  dif- 

~ [-1=9.     ference,  the  result  will  be  the  number  of 

common  differences;  thus  19  —  3  =  16  ; 
16  -J-  2  =  8.      Then,  as  the   number  of 
terms  must  be  1  more  than  the  number  of  common  differences,  8  -|- 
1  =  9  is  the  number  of  terms  in  the  series. 

Rule.  —  Divide  the  difference  of  the  extremes  by  the  common  differ- 
ence, and  the  quotient  increased  by  1  will  be  the  required  number  of 
terms. 

Examples. 

2.  A  man  going  a  journey  travelled  the  first  day  7  miles,  the 
last  day  51  miles,  and  each  day  increased  his  journey  by  4 
miles.     How  many  days  did  he  travel  ?  Ans.  12. 

3.  In  what  time  can  a  debt  be  discharged,  supposing  the  first 
week's  payment  to  be  S 1,  and  the  payment  of  every  succeed- 
ing week  to  increase  by  S  2,  till  the  last  payment  shall  be 
S  103  ?  Ans.  52  weeks. 

559.  To  find  the  sum  of  all  the  terms,  the  extremes  and 
number  of  terms  being  given. 

Ex.  1.  The  extremes  of  an  arithmetical  series  are  3  and 
19,  and  the  number  of  terms  9.  Required  the  sum  of  the 
series.  Ans.  99. 

operation.  In  an  arithmetical 

3-J-19  ooa  series  the  sum  of  the 

o  X  9  =  99j  Ans.  extremes  is  equal  to 

the  sum  of  two  terms 
Or,  3  +  19  =  22  ;  22  X  H  =  09,  Ans.  that  are  equally  dis- 
tant from  them,  or  to 
double  the  middle  term,  if  the  number  of  terms  be  odd.  Thus,  in  the 
series,  3,  5,  7,  9,  11,  13,  15,  17,  19,  the  sum  of  3  and  19  is  equal  to 
the  sum  of  5  and  17,  or  of  7  and  15,  and  is  double  the  middle  term, 
11.  The  reason  of  this  is  evident,  since  5  and  7  exceed  the  less  ex- 
treme by  the  same  quantities  by  which  17  and  15  are  respectively 
less  than  the  other  extreme. 
N       34 


398  PROGRESSION,   OR  SERIES. 

Hence,  in  this  latter  series,  it  is  evident  that,  if  each  terra  were 
made  11,  half  the  sum  of  the  extremes,  the  sum  of  the  whole  would 
remain  the  same ;  therefore  the  sum  of  the  series  must  equal  half  the 
sum  of  the  extremes  multiplied  by  the  number  of  terms,  or  the  sum 
of  the  extremes  multiplied  by  half  the  number  of  terms. 

Kule.  —  Multiply  half  the  sum  of  the  extremes  by  the  number  of 
terms.     Or, 

Multiply  the  sum  of  the  extremes  by  half  the  number  oftemxs. 

Examples. 

2.  If  the  least  term  of  a  series  of  numbers  in  arithmetical 
progression  be  4,  the  greatest  100,  and  the  number  of  terms  17, 
what  is  the  the  sum  of  the  terms  ?  Ans.  884. 

3.  Suppose  a  number  of  stones  were  laid  a  rod  distant  from 
each  other,  for  thirty  miles,  the  first  stone  being  a  rod  from  a 
basket.  What  distance  will  that  man  travel  who  gathers  them 
up  singly,  returning  with  them  one  by  one  to  the  basket  ? 

Ans.  288090  miles  2  rods. 

560»  To  find  the  sum  of  the  terms,  the  extremes  and  com- 
mon difference  being  given. 

Ex.  1.  If  the  two  extremes  are  3  and  19,  and  the  common 
difference  is  2,  what  is  the  sum  of  the  series  ? 

operation.  ^  nas  been  shown  (Art  558)  that, 

2Q 3  if  the  difference  of  the  extremes  be 

[-1  =  9;  divided  by  the   common  difference, 

2  the  quotient  will  be  the  number 
19  -|-  3  Q  ooa  °^  terms  *ess  *one*  Therefore  the 
2 ^                  '  Ans.  number  of  terms  less  one  will  be 

■   I"    ==  8  ;  and  8  -J- 1  will  equal  the 

number  of  terms.  It  has  also  been  shown  (Art.  559)  that,  if  the 
number  of  terms  be  multiplied  by  the  sum  of  the  extremes,  and  the 
product  divided  by  2,  the  quotient  will  be  the  sum  of  the  series; 

therefore  — j~  X  9  =  99,  the  answer  required. 

BULB,  — Divide  the  difference  of  the  extremes  by  the  common  differ- 
ence, and  to  the  quotient  add  1 ;  by  this  sum  multiply  half  the  sum  of  the 
extremes,  and  the  product  will  be  the  sum  required. 

Examples. 

2.  If  the  extremes  are  3  and  45,  and  the  common  difference 
2,  what  is  the  sum  of  the  series  ?  Ans.  528. 

3.  A  owes  B  a  certain  sum,  to  be  discharged  in  a  year,  by 


PROGRESSION,   OK   SERIES.  399 

paying  6  cents  the  first  week,  18  cents  the  second  week,  and 
thus  to  increase  every  week  by  12  cents,  till  the  last  payment 
should  be  $  6.18.     What  is  the  debt  ?  Ans.  $  162.24. 

561 .  To  find  one  of  the  extremes,  when  the  other  extreme 
and  the  number  and  sum  of  the  terms  are  given. 

Ex.  1.  If  3  be  the  first  term  of  a  series,  9  the  number  of 
terms,  and  99  the  sum  of  the  series,  what  is  the  last  term  ? 

operation.  It  has  been  shown  (Art.  559)  that,  if  the 

99  X  2  .  sum  of  the  extremes  be  multiplied  by  the 

=22,  Ans.     number  of  terms,  the  product  will  be  twice 

the  sum  of  the  series ;  therefore,  if  twice  the 
sum  of  the  series  be  divided  by  the  number  of  terms,  the  quotient 
will  be  the  sum  of  the  extremes.  If  from  this  we  subtract  the  given 
extreme,  the  remainder  must  be  the  other  extreme. 

Rule.  —  Divide  twice  the  sum  of  tlie  series  by  the  number  of  terms  ; 
from  the  quotient  take  the  given  term,  and  the  remainder  will  be  the  term 
required. 

Examples. 

2.  The  sum  of  a  series  of  ten  thousand  even  numbers  i3 
100010000,  and  the  last  term  of  the  series  is  20000.  Required 
the  first  term.  Ans.  2. 

3.  A  merchant,  being  indebted  to  22  creditors  $528,  ordered 
his  clerk  to  pay  the  first  %  3,  and  the  rest  sums  increasing  in 
arithmetical  progression.  What  is  the  difference  of  the  pay- 
ments, and  the  last  payment  ? 

Ans.  Difference  2  ;  last  payment  $  45. 

562.  To  find  any  number  of  arithmetical  mea?is,  the  ex- 
tremes and  the  number  of  terms  being  given. 

Ex.  1.  If  the  first  term  of  an  arithmetical  series  is  1,  the 
last  term  99,  and  the  number  of  terms  8,  what  are  the  second 
and  seventh  terms  of  the  series  ? 

Ans.  The  second  term,  15  ;  the  seventh,  85. 

We  find  the  common  differ- 

opekation.  ence,  14,  as  in  Art.  557,  the 

qq     __  -j  first  term,  1,  plus  the  common 

—  14 ;  difference,    14,   gives    15    for 

8  —  1  the  second  term,  and  the  last 

iiia  i^.qo         i/i     -  qz       term,  99,  minus  the  common 

1  ~r  i4:  —  I0  >  JJ  —  i4  -  b0-     difference,  14,  gives  the  sev- 
enth term. 


400  PROGRESSION,   OR   SERIES. 

Rule.  —  Find  the  common  difference,  which,  added  to  the  less  ex- 
treme, or  subtracted  from  the  greater,  will  give  one  mean.  From  that 
mean  derive  others  in  the  same  v:ay,  till  those  required  are  found. 

Examples. 

2.  The  extremes  of  a  series  are  4  and  49,  and  the  number 
of  terms  6.     Required  the  middle  two  terms. 

Ans.  22  and  31. 

3.  Insert  five  arithmetical  means  between  20  and  30. 

Ans.  21§,  23£,  25,  26§,  and  28£. 


GEOMETRICAL  PROGRESSION. 

563 •  Geometrical  Progression,  or  progression  by  quotients, 
is  a  series  of  numbers  that  increase  or  decrease  by  a  constant 
multiplier  or  divisor,  called  the  common  ratio. 

The  series  is  an  ascending  one,  when  each  term  after  the  first 
increases  by  a  constant  ratio  ;  and  a  descending  one,  when  each 
term  after  the  first  decreases  by  a  constant  ratio. 

Thus  2,  6,  18,  54,  162,  486  is  an  ascending  geometrical 
series;  and  64,  32,  16,  8,  4,  2  is  a  descending  geometrical 
series.  Of  the  former,  3  is  the  common  ratio,  and  of  the  lat- 
ter, 2. 

564 1  In  geometrical  progression  theirs*  term,  the  last  term, 
the  number  of  terms,  the  common  ratio,  and  the  sum  of  the  terms 
are  so  related  to  each  other,  that,  any  three  of  these  being 
given,  the  other  two  may  be  readily  determined. 

565 1  To  find  any  proposed  term,  one  of  the  extremes,  the 
ratio,  and  the  number  of  terms  being  given. 

Ex.  1.  If  the  first  term  of  a  geometrical  series  be  3,  the 
ratio  2,  and  the  number  of  terms  8,  what  is  the  last  term  ? 

Ans.  384. 

operation.  It  is  evident  that  the  successive 

3  X  27  =  3  X  128  =  384.     terms  are  the  result  of  repeated 

multiplications  by  the  ratio ;  thus 
the  second  term  must  be  the  product  of  the  first  term  by  the  ratio, 
the  third  term  the  product  of  the  second  term  by  the  ratio,  and  so 
on.     The  eighth,  or  last  term,  therefore,  must  be  the  result  of  seven 


PROGRESSION,   OR   SERIES.  401 

such  multiplications,  or  the  product  of  the  first  term,  3,  by  27,  or 

3  X   128  =  384. 

If  the  last  term  had  been  given  and  the  first  required,  the  process 
would  evidently  have  been  by  division,  since  every  less  term  is  the 
result  of  a  division  of  the  term  next  larger  by  the  ratio. 

Rule.  —  Raise  the  ratio  to  the  power  ivhose  index  is  one  less  than  the 
number  of  terms  ;  by  ichich  multiply  the  least  term  to  find  the  greatest, 
or  divide  the  greatest  to  find  the  least. 

Note  1.  —  When  the  ratio  requires  to  be  raised  to  a  high  power,  the  process 
may  be  abridged,  as  in  Art  516. 

Note  2.  —  The  rule  may  be  applied  in  computing  compound  interest,  the 
principal  being  the  first  term,  the  amount  of  one  dollar  for  one  year  the  ratio, 
the  time,  in  years,  one  less  than  the  number  of  terms,  and  the  amount  the  last 
term. 

Examples. 

2.  If  the  first  term  be  5,  and  the  ratio  3,  what  is  the  seventh 
term  ?  Ans.  3645. 

3.  If  the  series  be  72,  24,  8,  &c,  and  the  number  of  terms  6, 
what  is  the  last  term  ?  Ans.  /T. 

4.  If  the  larger  extreme  be  885735,  the  ratio  3,  and  the 
number  of  terms  12,  what  are  the  tenth  and  the  eleventh 
terms  ?  Ans.  15  and  45. 

5.  If  the  seventh  term  is  5,  and  the  ratio  -J,  what  is  the  first 
term  ?  Ans.  3645. 

6.  If  the  first  term  is  50,  the  ratio  1.06,  and  the  number  of 
terms  5,  what  is  the  last  term  ?  Ans.  63.123848. 

7.  If  I  were  to  buy  30  oxen,  giving  2  cents  for  the  first  ox, 

4  cents  for  the  second,  8  cents  for  the  third,  &c,  what  would  be 
the  price  of  the  last  ox  ?  Ans.  $  10737418.24. 

8.  What  is  the  amount  of  $  160.00  at  compound  interest  for 
6  years  ?  Ans.  $  226.96305796096. 

9.  What  is  the  amount  of  $  300.00  at  compound  interest  at 

5  per  cent,  for  8  years  ?  Ans.  %  443.236+. 

10.  What  is  the  amount  of  $  100.00  at  compound  interest  at 

6  per  cent,  for  30  years  ? 

Ans.  $  574.349117291325011626410633231080264584635- 
7252196069357387776. 

566.     To  find  the  sum  of  a  series,  the  first  term,  the  ratio, 
and  the  number  of  terms  being  given. 
34* 


402  PROGRESSION,   OR   SERIES 


Ex.  1.     If  the  first  term  be  1,  the  ratio  3,  and  the  number 
of  terms  5,  required  the  sum  of  the  terms.  Ans.  121. 

operation.  If  we  multiply  the  series  1,  3,  9,  27, 

(81   X  3)  —  1     __  -91      81  by  the  ratio  3,  we  shall  obtain  as  a 

3 1  ~~  second  series  3,  9,  27,  81,  243,  whose 

sum  is  three  times  the  sum  of  the  first  se- 
ries, and  the  difference  between  whose  sum  and  the  sum  of  the  first 
series  is  evidently  twice  the  sum  of  the  first  series.  Now  it  will  be 
observed  that  the  two  series  have  their  terms  alike,  with  the  excep- 
tion of  the  first  term  in  the  first  series,  and  the  last  in  the  second 
series.  We  have  then  only  to  subtract  the  first  term  in  the  first 
series  from  the  last  term  in  the  second,  and  the  remainder  is  twice 
the  sum  of  the  first  series ;  and  half  this  being  taken  gives  the  re- 
quired sum  of  the  series.  Therefore  the  sum  of  the  first  series  must 
be  242  -f-  2  =  121. 

Rule.  —  Find  the  last  term  as  in  Art.  565,  multiply  it  by  the  ratio, 
and  the  product  less  the  Jirst  term  divide  by  the  ratio  less  1 ;  the  result 
will  be  the  sum  of  the  series. 

Note  1.  —  If  the  ratio  is  less  than  a  unit,  the  product  of  the  last  term  mul- 
plied  by  the  ratio  must  be  subtracted  from  the  first  term,  and  the  remainder 
divided  by  unity  or  1  decreased  by  the  ratio. 

Note  2.  —  When  a  descending  series  is  continued  to  infinity,  it  becomes 
what  is  called  an  infinite  series,  whose  last  term  must  always  be  regarded  as  0, 
and  its  ratio  as  a  fraction.     To  find  the  sum  of  an  infinite  series,  — 

Divide  the  first  term  by  1  decreased  by  the  fraction  denoting  the  ratio,  and  the 
quotient  will  be  the  sum  required. 

This  process  furnishes  an  expeditious  way  of  finding  the  value  of  circulating 
decimals,  since  they  are  composed  of  numbers  in  geometrical  progression, 
whose  common  ratios  are  Jj,  ^,  iqCC,  &c.  according  to  the  number  of  fac- 
tors contained  in  the  repetend.  Thus,  .3333,  &c.  represents  the  geometrical 
series  j0,  jgg,  i^q,  &c.  whose  first  term  is  fa  and  common  ratio  jq. 

Examples. 

2.  The  first  term  of  a  series  is  5,  the  ratio  §,  and  the  number 
of  terms  6  ;  required  the  sum  of  the  series. 

(i)5  x  5  =  n%;  m  x  §  -  m ;  s  -  m  =  Wj 

3.  Find  the  value  of  the  circulating  decimal  .232323,  &c. 

t2o%  -*-  (1  —  To-o-)  =  ih  Ans. 

4.  What  is  the  sum  of  the  series  4,  1,  £,  T\,  &c.,  continued 
to  an  infinite  number  of  terms  ? 

1-4-  (1  _  £)  =  5i   Ans. 

5.  If  the  first  term  is  50,  the  ratio  1.06,  and  the  number  of 
terms  4,  what  is  the  sum  of  the  series?  Ans.  218.7308. 


PROGRESSION,   OR   SERIES.  403 

6.  A  gentleman  offered  a  house  for  sale  on  the  following 
terms  ;  that  for  the  first  door  he  should  charge  10  cents,  for  the 
second  20  cents,  for  the  third  40  cents,  and  so  on  in  a  geometri- 
cal ratio.  If  there  were  40  doors,  what  was  the  price  of  the 
house  ?  Ans.  $  109951162777.50. 

7.  If  the  series  f ,  ^,  -|,  y1^,  ^,  &c.  were  carried  to  infinity, 
what  would  be  its  sum  ?  Ans.  1£. 

8.  A  gentleman  deposited  annually  $  10  in  a  bank,  from  the 
time  his  son  was  born  until  he  was  20  years  old.  Required  the 
amount  of  the  21  deposits  at  6  per  cent.,  compound  interest, 
when  his  son  was  21  years  old?  Ans.  $423.92. 

9.  Find  the  value  of  .008497133,  &c.,  continued  to  in- 
finity. Ans.  sf  h- 

10.  If  a  body  be  put  in  motion  by  a  force  which  moves  it  10 
miles  in  the  first  portion  of  time,  9  miles  in  the  second  equal 
portion,  and  so,  in  the  ratio  of  T9^-,  for  ever,  how  many  miles 
will  it  pass  over  ?  Ans.  100  miles. 

567 1  To  find  the  ratio,  the  extremes  and  number  of  terms 
being  given. 

Ex.  1.  If  the  extremes  of  a  series  are  3  and  192,  and  the 
number  of  terms  7,  what  is  the  ratio  ? 

operation^  It  has  been  shown  (Art. 

192  -i-  3  =  64 ;  /</64  sss  2,  Ans.     565)  that  the  last  term  of  a 

geometrical  series  is  equal  to 
the  product  of  the  ratio  by  the  first  term  raised  to  a  power  whose  in- 
dex is  one  less  than  the  number  of  terms ;  hence  the  ratio  must  equal 
the  root  of  the  quotient  of  the  last  term  by  the  first  whose  index  is 
one  less  than  the  number  of  terms. 

Rule.  —  Divide  the  last  extreme  by  the  first,  and  extract  that  root  of 
the  quotient  whose  index  is  one  less  than  the  number  of  terms. 

Note.  —  When  the  sum  of  the  series  and  the  extremes  are  given,  the  ratio 
may  be  found  by  dividing  the  sum  of  all  the  terms  except  the  first,  by  the  sum  of 
all  the  terms  except  the  last. 

Examples. 

2.  If  the  last  term  of  a  series  is  1,  the  largest  term  512,  and 
the  number  of  terms  10  ;  what  is  the  ratio  ?  Ans.  2. 

3.  If  the  extremes  are  5  and  885735,  and  the  sum  of  the  se- 
ries 1328600,  what  is  the  ratio?  Ans.  3. 

4.  What  debt  can  be  discharged  in  a  year  by  monthly  pay- 


404  PROGRESSION,   OR   SERIES. 

merits,  in  geometrical  progression,  of  which  the  first  payment  is 
$  1,  and  the  last  $  2048 ;  and  what  will  be  the  ratio  of  the  se- 
ries ?  Ans.  Ratio,  2 ;  debt,  $  4095. 

568.  To  insert  any  number  of  geometrical  means,  or  mean 
proportionals,  between  two  given  numbers. 

Ex.  1.     Insert  three  geometrical  means  between  4  and  324. 

Ans.  12,  36,  and  108. 

operation.  Since    tlio 

09A      •     A    —   Q1    .     J/tf  1    _   Q  SerIeS  wIU   in" 

6Z4:  -r-4  —  81  ;    A/bl  —  6.  cju(le     ^j^ 

4  X  3  =  12;  12  X  3  =  36;  36  X  3  =  108.     the    inserted 

terms,  the  two 
extremes,  the  number  of  terms  will  be  5.  Then,  having  the  number 
of  terms  and  the  extremes,  we  find  the  ratio,  as  in  the  last  article,  to 
be  3 ;  and  by  multiplying  the  first  term  by  the  ratio,  we  obtain  the 
first  of  the  terms  to  be  inserted.  That  term  multiplied  by  the  ratio 
gives  the  next,  and  that  multiplied  by  the  ratio  gives  the  other  re- 
quired mean. 

Rule. —  Take  tlie  two  given  numbers  as  the  extremes  of  a  geometri- 
cal series,  and  consider  the  number  of  terms  in  the  series  greater  by  two 
than  the  required  number  of  means.  Then  find  the  ratio,  as  in  Art.  5G7, 
and  the  product  of  the  ratio  and  the  first  extreme  icill  give  one  of  the 
means,  and  the  product  of  this  mean  and  the  ratio  will  give  another,  and 
so  on. 

Note.  —  When  only  a  single  mean  is  required  to  be  inserted,  it  may  be 
found  as  in  Art.  550;  when  only  two,  as  in  Art.  551. 

Examples. 

2.  Insert  three  geometrical  means  between  \  and  128. 

Ans.  2,  8,  and  32. 

3.  Required  five  mean  proportionals  between  the  numbers  3 
and  2187.  Ans.  9,  27,  81,  243,  and  729. 

569.  To  find  the  number  of  terms,  the  extremes  and  ratio 
being  given. 

Ex.  1.  If  the  extremes  are  5  and  3645,  and  the  ratio  3, 
what  is  the  number  of  terms  ? 

operation.  By  Art.  565  it  is  seen  that  the  ratio 

3645  -r-  729  ;  36  =  729.     raised  to  the  power  whose  index  is  one 
q    i    ^  =  7   Ans.  ^ess  than  the  number  of  terms,  and 

■^  multiplied  by  the  least  term,  equals  the 

largest  term ;  hence,  the  largest  term  divided  by  the  least  term  will 
equal  a  power  of  the  ratio  whose  index  is  one  less  than  the  number 
of  terms. 


ANNUITIES.  405 

Rule.  — Divide  the  largest  term  by  the  least;  involve  the  ratio  to  a 
power  equal  to  the  quotient;  and  the  index  of  that  power,  increased  by 
1,  will  be  the  number  of  terms. 

Examples. 

2.  If  the  extremes  are  5  and  20480,  and  the  ratio  4,  what  is 
the  number  of  terms  ?  Ans.  7. 

3.  In  what  time  will  a  certain  debt  be  discharged  by  monthly 
payments  in  geometrical  progression,  if  the  first  and  last  pay- 
ments are  $  1  and  $  2048,  and  the  ratio  2  ? 

Ans.  In  12  months. 


ANNUITIES. 


570.  Annuities  are  fixed  sums  of  money  payable  at  the 
ends  of  equal  periods  of  time,  such  as  years,  or  half-years. 

Annuities  in  perpetuity  are  such  as  continue  for  ever. 

Annuities  certain  are  such  as  commence  at  a  fixed  time,  and 
continue  for  a  certain  number  of  years. 

Annuities  contingent  are  those  whose  commencement  or  con- 
tinuance, or  both,  depend  on  some  contingent  event,  as  the 
death  of  one  or  more  individuals. 

Annuities  deferred,  or  in  reversion,  are  such  as  do  not  com- 
mence till  after  a  fixed  number  of  years,  or  till  after  some  par- 
ticular event  has  taken  place. 

571.  An  annuity  forborne,  or  in  arrears,  is  one  whose 
periodical  payments,  instead  of  being  paid  when  due,  have 
been  allowed  to  accumulate. 

572.  The  amount  of  an  annuity  at  compound  interest,  at 
any  time,  is  the  sum  to  which  it  will  amount,  supposing  it  to 
have  been  improved  at  compound  interest  during  the  interven- 
ing period. 

573.  The  present  value  of  an  annuity  at  compound  interest, 
for  any  given  period,  is  the  sum  of  the  present  values  of  all  the 
payments  of  that  annuity. 


406 


ANNUITIES. 


TABLE, 


SNOWING  THE  AMOUNT  OF  AN  ANNUITY  OF  ONE  DOLLAR  TER  ANNUM, 
IMPROVED  AT  COMPOUND  INTEREST  FOR  ANY  NUMBER  OF  YEARS  NOT 
EXCEEDING  FIFTY. 


t 

•A 

i 

3  per  cent. 

3|  per  cent. 

4  per  cent. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

1.000  ooo 

!  1.000  ooc 

>  1.000  ooc 

1.000  000 

1.000  000 

1.000  000 

2 

2.030  000 

!  2.035  000 

2.040  000 

2.050  000 

2.060  000 

2.070  000 

3 

3.090  900 

3.106  225 

3.121  600 

3.152  500 

3.183  600 

3.214  900 

4 

4.183  627 

4.214  943 

4.246  464 

4.310  125 

4.374  616 

4.439  943 

5 

5.309  136 

5.362  466 

5.416  323 

5.525  631 

5.637  093 

5.750  739 

6 

6.468  410 

6.550  152 

6.632  975 

6.801  913 

6.975  319 

7.153  291 

7 

7.662  462 

7.779  408   7.898  294 

8.142  008 

8.393  838 

8.654  021 

8 

8.892  336 

9.051  687 

9.214  226 

9.549  109 

9.897  468 

10.259  803 

9 

10.159  106 

10.368  496 

10.582  795 

11.026  564 

11.491  316 

11.977  989 

10 

11.463  879 

11.731  393 

12.006  107 

12.577  893 

13.180  795 

13.816  448 

11 

12.807  796 

13.141  992 

13.486  351 

14.206  787 

14.971  643 

15.783  599 

12 

14.192  030 

14.601  962 

15.025  806 

15.917  127 

16.869  941 

17.888  451 

13 

15.617  790 

16.113  030 

16.626  838 

17.712  983 

18.882  138 

20.140  643 

14 

17.086  324 

17.676  986 

18.291  911 

19.598  632 

21.015  066 

22.550  488 

15 

18.598  914 

19.295  681 

20.023  588 

21.578  564 

23.275  970 

25.129  022 

16 

20.156  881 

20.971  030 

21.824  531 

23.657  492 

25.670  528 

27.888  054 

17 

21.761  588 

22.705  016 

23.697  512 

25.840  366 

28.212  880 

30.840  21? 

18 

23.414  435 

24.499  691 

25.645  413 

28.132  385 

30.905  653 

33.999  033 

19 

25.116  868 

26.357  180 

27.671  229 

30.539  004 

33.759  992 

37.378  965 

20 

26.870  374 

28.279  682 

29.778  079 

33.065  954 

36.785  591 

40.995  492 

21 

28.676  486 

30.269  471 

31.969  202 

35.719  252 

39.992  727 

44.865  177 

22 

30.536  780 

32.328  902 

34.247  970 

38.505  214 

43.392  290 

49.005  739 

23 

32.452  884 

34.460  414 

36.617  889 

41.430  475 

46.995  828 

53.436  141 

24 

34.426  470 

36.666  528 

39.082  604 

44.501  999 

50.815  577 

58.176  671 

25 

36.459  264 

38.949  857 

41.645  908 

47.727  099 

54.864  512 

63.249  030 

26 

38.553  042 

41.313  102 

44.311  745 

61.113  454 

59.156  383 

68.676  470 

27 

40.709  634 

42.759  060 

47.084  214 

54.669  126 

63.705  766 

74.483  823 

28 

42.930  923 

46.290  627 

49.967  583 

58.402  583 

68.528  112 

80.697  691 

29 

45.218  850 

48.910  799 

52.966  286 

62.322  712 

73.639  798 

87.346  529 

30 

47.575  416 

51.622  677 

56.084  938 

66.438  848 

79.058  186 

94.460  786 

31 

50.002  678 

54.429  471  59.328  335 

70.760  790 

84.801  677 

102.073  041 

32 

52.502  759 

57.334  502  62.701  469 

75.298  829 

90.889  778, 

110.218  154 

33 

55.077  841 

60.341  210:  66.209  527 

80.063  771 

97.343  165! 

118.933  425 

34 

57.730  177 

63.453  152  69.857  909 

85.066  959: 

104.183  755' 

128.258  765 

35 

60.462  082 

66.674  013  73.652  225 

90.320  307i 

111.434  780; 

138.236  878 

36 

63.271  944 

70.007  603|  77.598  314 

95.836  323 

119.120  867i 

148.913  460 

37 

66.174  223 

73.457  869  81.702  246 

101.628  139 

127.268  119 

160.337  400 

38 

69.159  449 

77.028  895  85.970  336! 

107.709  546; 

135.904  206: 

172.561  020 

39 

72.234  233 

80.724  906  90.409  150| 

114.095  023! 

145.058  458; 

185.640  292 

40 

75.401  260 

84.550  278|  95.025  516 

120.799  774! 

154.761  966; 

199.635  112 

41 

78.663  298 

88.509  537|  99.826  536' 

127.839  763: 

165.047  684 

214.609  570 

42 

82.023  196 

92.607  371  104.819  598, 

135.231  751! 

175.950  645 

230.632  240 

43 

85.483  892 

96.848  629  110.012  382; 

142.993  339 

187.507  5771 

247.776  496 

44 

89.048  409 

101.238  331  115.412  877) 

151.143  006,'l99.758  032 

266.120  851 

45 

92.719  861 

105.781  673  121.029  392 

159.700  156  212.743  514, 

285.749  311 

46 

96.501  4571 

110.484  031  126.870  568 

168.685  164226.508  125 

306.751  763 

47 

100.396  601 

115.350  973  132.945  390 

178.119  422  241.098  612] 

329.224  386 

48 

104.408  396 

120.388  297 

139.263  206 

188.025  393i256.564  529 

353.270  093 

49 

108.540  648 

125.601  846 

145.833  734 

198.426  663j272.958  401 

378.999  000 

50 

112.796  867  130.999  910 

152.667  084  209.347  9761290.335  905  406.528  929 

ANNUITIES, 


407 


TABLE, 


SHOWING  THE  PRESENT  WORTH  OF  AN  ANNUITY  OF  ONE  DOLLAR  PER 
ANNUM,  TO  CONTINUE  FOR  ANY  NUMBER  OF  YEARS  NOT  EXCEEDING 
FIFTY. 


g 

H 
1 

3  per  cent. 

3£  per  cent. 

4  per  cent. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

I 

l 

0.970  874 

0.966  184 

0.961  538 

0.952 

381 

0.943 

396 

0.934 

579 

2' 

1.913  470 

1.899  694 

1.886  095 

1.859 

410 

1.833 

393 

1.808 

017 

2 

3 

2.828  611 

2.801  637 

2.775  091 

2.723 

248 

2.673 

012 

2.624 

314 

3 

4 

3.717  098 

3.673  079 

3.629  895 

3.545 

951 

3.465 

106 

3.387 

209 

4 

5 

4.579  707 

4.515  052 

4.451  822 

4.329 

477 

4.212 

364 

4.100 

195 

5 

6 

5.417  191 

5.328  553 

5.242  137 

5.075 

692 

4.917 

324 

4.766 

537 

6 

7 

6.230  283 

6.114  544 

6.002  055 

5.786 

373 

5.582 

381 

5.389 

286 

7 

8 

7.019  692 

6.873  956 

6.732  745 

6.463 

213 

6.209 

744 

5.971 

295 

8 

9 

7.786  109 

7.607  687 

7.435  332 

7.107 

822 

6.801 

692 

6.515 

228 

9 

10 

8.530  203 

8.316  605 

8.110  896 

7.721 

735 

7.360 

087 

7.023 

577 

10 

11 

9.252  624 

9.001  551 

8.760  477 

8.306 

414 

7.886 

875 

7.498 

669 

11 

12 

9.954  004 

9.663  334 

9.385  074 

8.863 

252 

8.383 

844 

7.942 

671 

12 

13 

10.634  955 

10.302  738 

9.985  648 

9.393 

573 

8.852 

683 

8.357 

635 

13 

14 

11.296  073 

10.920  520 

10.563  123 

9.898 

641 

9.294 

984 

8.745 

452 

14 

15 

11.937  935 

11.517  411 

11.118  387 

10.379 

658 

9.712 

249 

9.107 

898 

15 

16 

12.561  102 

12.094  117 

11.652  296 

10.837 

770 

10.105 

895 

9.446 

632 

16 

17 

13.106  118 

12.651  321 

12.165  669 

11.274 

066 

10.477 

260 

9.763 

206 

17 

18 

13.753  513 

13.189  682 

12.659  297 

11.689 

587 

10.827 

603 

10.059 

070 

18 

19 

14.323  799 

13.709  837 

13.133  939 

12.085 

321 

11.158 

116 

10.335 

578 

19 

20 

14.877  475 

14.212  403 

13.590  326 

12.462 

210 

11.469 

421 

10.593 

997 

20 

21 

15.415  024 

14.697  974 

14.029  160 

12.821 

153 

11.764 

077 

10.835 

527 

21 

22 

15.936  917 

15.167  125 

14.451  115 

13.163 

003 

12.041 

582 

11.061 

241 

22 

23 

16.443  608 

15.620  410 

14.856  842 

13.488 

574 

12.303 

379 

11.272 

187 

23 

24 

16.935  542 

16.058  368 

15.246  963 

13.798 

642 

12.550 

358 

11.469 

334 

24 

25 

17.413  148 

16.481  515 

15.622  080 

14.093 

945 

12.783 

356 

11.653 

583 

25 

26 

17.876  842 

16.890  352 

15.982  769 

14.275 

185 

13.003 

166 

11.825 

779 

26 

27 

18.327  031 

17.285  365 

16.329  586 

14.643 

034 

13.210 

534 

11.986 

709 

27 

28 

18.764  108 

17.667  019 

16.663  063 

14.898 

127 

13.406 

164 

12.137 

111 

28 

29 

19.188  455 

18.035  767 

16.983  715 

15.141 

074 

13.590 

721 

12.277 

674 

29 

30 

19.600  441 

18.392  045 

17-292  033 

15.372 

451 

13.764 

831 

12.409 

041 

30 

31 

20.000  428 

18.736  276 

17.588  494 

15.592 

811 

13.929 

086 

12.531 

814 

31 

32 

20.338  766 

19.068  865 

17.873  552 

15.802 

677 

14.084 

043 

12.646 

555 

32 

33 

20.765  792 

19.390  208 

18.147  646 

16.002 

549 

14.230 

230 

12.753 

790 

33 

34 

21.131  837 

19.700  684 

18.411  198 

16.192 

204 

14.368 

141 

12.854 

009 

34 

35 

21.487  220 

20.000  661 

18.664  613 

16.374 

194 

14.498 

246 

12.947 

672 

35 

36 

21.832  252 

20.290  494 

18.908  282 

16.546 

852 

14.620 

987 

13.035 

208 

36 

37 

22.167  235 

20.570  525 

19.142  579 

16.711 

287 

14.736 

780 

13.117 

0171  37 

38 

22.492  462 

20.841  087 

19.367  864 

16.867 

893 

14.846 

019 

13.193 

473'  38 

39 

22.808  215 

21.102  500 

19.584  485 

17.017 

041 

14.949 

075 

13.264 

928|  39 

40 

23.114  772 

21.355  072 

19.792  774 

17.159 

086 

15.046 

297 

13.331 

709|  40 

41 

23.412  400 

21.599  104 

19.993  052 

17.294 

36S 

15.138 

016 

13.394 

120  41 

42 

23.701  359 

21.834  883 

20.185  627 

17.423 

208 

15.224 

543 

13.452 

449 

42 

43 

23.981  902 

22.062  689 

20.370  795 

17.545 

912 

15.306 

173 

13.506 

962 

43 

44 

24.254  274 

22.282  791 

20.548  841 

17.662 

773 

15.383 

182 

13.557 

908 

44 

45 

24.518  713 

22.495  450 

20.720  040 

17.774 

070 

15.455 

832 

13.605 

522 

45 

40 

24.775  449 

22.700  918 

20.884  654 

17.880 

067 

15.524 

370 

13.650 

020 

46 

47 

25.024  708 

22.899  438 

21.042  936 

17.981 

016 

15.589 

028 

13.691 

608 

47 

48 

25.266  707 

23.091  244 

21.195  131 

18.077 

158 

15.650 

027 

13.730 

474 

48 

49 

25.501  657 

23.276  564 

21.341  472 

18.168 

722 

15.707 

572  13.766 

799 

49 

50 

25.729  764 

23.455  618 

21.482  185  18.255 

925 

15.761 

861113.800 

746  50 

408  ANNUITIES. 

574.     To  find  the  amount  of  an  annuity,  at  compound  in- 
terest, forborne,  or  in  arrears,  for  any  number  of  years. 

Ex.  1.     What  will  an  annuity  of  $  60,  unpaid,  or  in  arrears, 
4  years,  amount  to,  at  6  per  cent,  compound  interest  ? 

Ans.  $262,476. 

The  amounts  of  the  several 

operation.  payments  form  a  geometrical 

1.064 —  1  series,  of  which  the  annuity  is 

J^q6  2   X  60  =  262.4/6.     tiie  first  term,  the  amount  of 

'  $  1  for  one  year  the  ratio,  the 

Or,  4.374616  X  GO  =  262.476.     years  the  number  of  the  terms, 

and  the  amount  required  is  the 
sum  of  the  series.     Hence, 

Rule.  —  Find  the  sum  of  the  series  as  in  geometrical  progression. 

0r' 

Multiply  the  amount  of  $  1  for  the  given  time,  found  in  the  table,  by 

the  annuity,  and  the  product  icdl  be  the  required  amount. 

Note.  —  The  amount  of  an  annuity  at  simple  interest  corresponds  to  the 
sum  of  an  arithmetical  series,  of  which  the  annuity  is  the  first  term,  the  interest 
on  the  annuity  for  one  term  the  common  difference,  and  the  time  in  years  the 
number  of  terms. 

2.  What  will  an  annuity  of  $  500  amount  to  for  5  years,  at 
6  per  cent,  compound  interest?  Ans.  $2818.546. 

3.  What  is  the  amount  of  an  annuity  of  $  80,  unpaid,  or  in 
arrears,  for  9  years,  at  5  per  cent,  compound  interest  ? 

Ans.  $882,125. 

4.  What  is  the  amount  of  an  annuity  of  $  1000,  forborne  for  - 
15  years,  at  3£  per  cent,  compound  interest  ? 

Ans.  $19295.68. 

5.  What  will  an  annuity  of  $  30,  payable  semiannually, 
amount  to,  in  arrears  for  3  years,  at  7  per  cent,  compound  in- 
terest ? 

6.  Suppose  a  salary  of  $  600  per  year,  payable  quarterly,  to 
remain  unpaid  5J-  years;  to  what  sum  will  it  amount,  at  6 
per  cent,  compound  interest  ?  .     Ans.  $  5404.295. 

575.  To  find  the  present  worth  of  an  annuity,  at  compound 
interest. 

Ex.  1.  What  is  the  present  worth  of  an  annuity  of  $  60,  to 
be  continued  4  years,  at  6  per  cent,  compound  interest  ? 

Ans.  $  207.906. 


ANNUITIES.  409 

operation.  The    present    worth    re- 

$  3.465106  X  60  =  $  207.906-j-.  quired  evidently  may  be  ob- 
tained by  finding  the  amount 
of  the  given  annuity,  by  the  last  articles,  and  then  finding  in  the 
usual  way  the  present  worth  of  that  amount.  A  more  expeditious 
method,  however,  is  to  find,  in  the  table,  the  present  worth  of  an  an- 
nuity of  $  1  for  the  given  time  and  rate,  and  take  that  sum  as  many 
times  as  there  are  dollars  in  the  given  annuity,  as  in  the  operation. 

Rule.  —  Multiply  the  present  worth  of  an  annuity  of  $1  for  the 
given  time  and  rate  by  the  number  denoting  the  given  annuity. 

2.  What  is  the  present  worth  of  an  annuity  of  $100,  for  9 
years,  at  6  per  cent.  ?  Ans.  $  680.169. 

3.  What  is  the  present  worth  of  an  annuity  of  $  200,  for  7 
years,  at  5  per  cent.  ?  Ans.  $  1157.27. 

4.  Eequired  the  present  worth  of  an  annuity  of  $  500,  to 
continue  40  years,  at  7  per  cent. 

5.  A  gentleman  wishes  to  purchase  an  annuity,  which  shall 
afford  him,  at  6  per  cent,  compound  interest,  $  500  a  year,  for 
ten  years.  What  sum  must  he  deposit  in  the  annuity  office  to 
produce  it  ?  Ans.  $  3680.04. 

6.  If  a  widow  be  entitled  to  $  160  a  year,  payable  semi- 
annually, from  a  fund,  for  8  years,  what  is  its  value  at  present, 
at  6  per  cent,  compound  interest?  Ans.  $  1004.88. 

576.     To  find  the  present  worth  of  an  annuity  in  perpetuity. 

Ex.  1.  What  is  the  present  worth  of  a  perpetual  lease, 
which  yields  an  income  of  $  600,  the  rate  of  interest  being  that 
of  6  per  cent.  ?  Ans.  $  10000. 

The  question  is  evidently  the  same 

operation.  as  one  requiring  what  principal  in  one 

$  600  -f-  .06  =  $  10000.     year,  at  6  per  cent,  interest,  will  yield 

$600. 

Rule.  —  Divide  the  given  annuity  by  the  number  denoting  the  inter- 
est of%\  for  one  year. 

Note.  —  When  the  annuity  is  payable  quarterly,  semiannually,  or  in  any 
other  periods  less  than  a  whole  year,  the  annuity  must  be  increased  by  the  in- 
terest which  may  thus  accrue  on  the  parts  of  the  annuity  payable  before  the 
end  of  the  year,  before  dividing  by  the  interest  of  $  1  for  one  year. 

2.  A  ground  rent  in  the  city  of  Philadelphia  yields  an  an- 
nual income  of  $963,  at  6  per  cent,  interest.  What  is  the 
value  of  the  estate  ?  Ans.  $  16050. 

N     35 


410  ANNUITIES. 

3.  "What  is  the  present  value  of  a  perpetual  lease,  yielding 
an  income  of  $  6335,  interest  being  at  7  per  cent. 

Ans.  $  90500. 

4.  What  sum  should  be  paid  for  a  perpetual  annuity  of 
$  1200,  payable  semiannually,  interest  being  at  5  per  cent.  ? 

Ans.  $  24000. 

577.  To  find  the  'present  worth  of  an  annuity  in  rever- 
sion. 

Ex.  1.  What  is  the  present  worth  of  an  annuity  of  $  300,  to 
commence  in  3  years,  and  to  continue  5  years,  allowing  com- 
pound interest  at  6  per  cent.?  Ans.  $  1061.03. 

operation.  The  present  wort li 

$6.209794  — $2.673012  =  $3.536772;  of  an  annuity  of  Si, 
$  3.53677  X  300  =  $  1061.03.  .  at  G  per  cent.,  com- 
mencing at  once, 
and  continuing  till  the  termination  of  the  annuity,  or  for  3  -f-  5  =  8 
years,  is  $  6.209794  ;  and  the  present  worth  of  the  same  annuity  up 
to  the  time  of  the  commencement  of  the  reversion  is  S  2.673012. 
The  difference  of  these  present  worths  multiplied  by  the  number  of 
dollars  in  the  given  annuity  is  the  present  worth  of  the  reversion. 

Rule.  —  Find  the  present  worth  of  an  annuity  of  $  1,  commencing 
immediately  and  continuing  till  the  reversion  commences,  and  also  till 
the  reversion  terminates  ;  and  multiply  the  difference  of  these  present 
worths  by  the  number  of  dollars  in  the  given  annuity.  The  result  will 
be  the  present  worth  required. 

2.  The  reversion  of  a  lease  of  $  350  per  annum,  to  continue 
1 1  years,  which  commences  9  years  hence,  is  to  be  sold.  What 
is  its  worth,  allowing  the  purchaser  6  per  cent,  per  annum  for 
his  ready  money  ?  Ans.  $  1633.70. 

3.  A  father  presents  to  his  daughter,  for  8  years,  a  rental  of 
$  70  per  annum,  payable  yearly,  and  its  reversion  for  the  12 
years  succeeding  to  his  son.  What  is  the  present  value  of  the 
gift  to  his  son,  allowing  4  per  cent,  compound  interest  ? 

Ans.  $  480.03. 

4.  What  is  the  present  worth  of  the  reversion  of  a  perpetuity 
of  $  240  per  annum,  payable  yearly,  but  not  to  come  into  pos- 
session till  the  expiration  of  100  years,  compound  interest  being 
allowed  at  6  per  cent.  ?  Ans.  $  11.78. 

578.  To  find  the  annuity,  the  present  worth,  time,  and  rate 
being  given. 


PERMUTATIONS   AND   COMBINATIONS.  411 

Ex.  1.  What  annuity,  continued  for  4  years,  at  6  per  cent, 
compound  interest,  will  amount  to  a  debt  of  $  207.90  ? 

Ans.  $  GO. 

The  present  value  represented  by 

operation.  the   debt,   divided  by   the   present 

$  207.90  -r-  3.465  =  $  60.     worth  of  $  1  for  the  given  time  and 

rate,  gives  the  annuity  required. 

Rule.  —  Divide  the  given  present  worth  by  the  present  worth  of  an 
annuity  of$  1  for  the  given  time  and  rate,  and  the  result  will  be  the  an- 
nuity required. 

Note.  —  When  the  amount  of  an  annuity,  the  time  and  rate,  are  given,  the 
annuity  may  be  found  by  dividing  the  given  amount  by  the  amount  of  $  1  for  the 
given  time  and  rate. 

2.  The  present  value  of  an  annuity,  to  be  continued  10 
years,  at  6  per  cent,  compound  interest,  payable  annually,  is 
$  3680.04 ;  required  the  annuity.  Ans.  $  500. 

3.  An  annuity,  remaining  unpaid  for  9  years,  at  5  per  cent, 
compound  interest,  amounted  to  $  882.125  ;  what  was  the  an- 
nuity ? 

4.  A  yearly  pension  which  has  been  forborne  for  6  years,  at 
6  per  cent,  amounts  to  $  279  ;  what  was  the  pension  ? 

Ans.  $  40. 


PERMUTATIONS  AND  COMBINATIONS. 

579.  Permutation  is  the  process  of  finding  the  number  of 
changes  that  can  be  made  in  the  arrangement  of  any  given 
number  of  things. 

580.  Combination  shows  how  often  a  less  number  of 
things  combined  can  be  taken  out  of  a  greater,  without  respect 
to  their  order. 

581 .  To  find  the  number  of  changes  that  can  be  made  with 
any  given  number  of  things,  taken  all  at  once. 

Ex.  1.  How  many  changes  of  order  do  the  first  three  letters 
of  the  alphabet  admit  of?  Ans.  6. 


412  x  PERMUTATIONS    AND    COMBINATIONS. 

opeiiation.  By  trial  we  shall  find  that  two  are  all  the 

1  X  2  X  ^  =  6.  possible  permutations  that  can  be  made  of  the 
first  two  letters  of  the  alphabet ;  as,  a  b  and 
b  a.  If  we  take  an  additional  letter,  6  are  all  the  possible  permuta- 
tions; as,  a  be,  acb,  bca,bac,  cab,  cba.  Now,  the  same  result  may 
be  obtained,  in  the  case  of  the  two  letters,  by  multiplying  together  the 
first  two  digits,  and  in  case  of  the  three  letters  by  multiplying  to- 
gether the  first  three  digits,  as  in  the  operation. 

Rule.  — Multiply  together  all  the  terms  of  the  natural  series  of  num- 
bers, from  1  up  to  the  given  number,  inclusive,  and  the  product  will  be 
the  number  required. 

2.  How  many  changes  may  be  rung  on  6  bells  ? 

Ans.  720  changes. 

3.  For  how  many  days  can  10  persons  be  placed  in  a  differ- 
ent position  at  dinner  ? 

4.  How  many  changes  may  be  rung  on  12  bells,  and  how 
long  would  they  be  in  ringing,  supposing  10  changes  to  be  rung 
in  one  minute,  and  the  year  to  consist  of  365  days  5  hours  and 
49  minutes?  Ans.  479001600,  and  91y.  2Gd.  22h.  41m. 

5.  How  many  changes  do  the  letters  of  the  alphabet  admit 
of?  Ans.  403291461126605635584000000. 

582.  To  find  how  many  changes  may  be  made  by  taking 
each  time  any  number  of  different  things  less  than  all. 

Ex.  1.  How  many  sets  of  4  letters  each  may  be  formed 
out  of  8  different  letters  ?  Ans.  1680. 

operation.  Ifc    is    evident    that 

8  X  (8-1)  X  (8-2)  X  (8-3)     S^bJSiS 
=  8X7X6X5  =  1680.  before  each  of  the  oth- 

ers ;  therefore  2  out  of 
the  8  letters  admit  of  8  X  7  permutations.  By  taking  3  out  of  the 
8  letters,  the  third  letter  can  be  arranged  as  the  first,  second,  and 
third,  in  each  of  the  same  permutations,  giving  8X7x6  permuta- 
tions. In  like  manner  for  4  out  of  8,  we  obtain  8X7x6x5  per- 
mutations. 

Rule.  —  Take  a  series  of  numbers,  beginning  with  the  number  of 
things  given,  and  decreasing  by  1,  until  the  number  of  terms  equals  the 
number  of  things  to  be  taken  at  a  time,  and  the  product  of  all  the  terms 
icill  be  the  answer  required. 

2.  How  many  changes  can  be  rung  with  4  bells  out  of  6  ? 

Ans.  360. 


ANALYSIS  BY  POSITION.  413 

3.  How  many  words  can  be  made  out  of  the  26  letters  of  the 
alphabet,  6  being  taken  at  once?  Ans.  165765600. 

583.  To  find  the  number  of  combinations  that  can  be 
formed  from  a  given  number  of  different  things,  taken  a  given 
number  at  a  time. 

Ex.  1.  How  many  combinations  can  be  made  of  3  letters 
out  of  4,  the  letters  all  being  different  ?  Ans.  4. 

operation.  We  find  the  number  of  permuta- 

4  X  &  X  $  tions  which  may  be  made  by  taking  3 

— -   =  y   =  4.       out  of  4  letters,  as  in  Art.  582,  and  the 

1X^X3  number  of  permutations  by  the  three 

letters  all  at  once,  as  by  Art.  581, 
and,  dividing  the  first  by  the  latter,  obtain  the  number  of  combina- 
tions required. 

Rule.  —  Take  tJie  series,  1,  2,  3,  4,  5,  fyc,  up  to  ilie  less  number  of 
things,  and  find  the  product  of  the  terms,  Take  also  a  series  of  num- 
bers beginning  with  the  greater  number  of  things,  and  decreasing  by  1 
until  the  number  of  terms  equals  the  less  number  of  things,  and  find  the 
product  of  the  terms.  The  latter  result  divided  by  the  former  will  give 
the  number  required. 

2.  How  many  combinations  can  be  made  of  7  letters  out  of 
10,  the  letters  all  being  different  ?  Ans.  120. 

3.  A  successful  general,  being  asked  what  reward  would  sat- 
isfy him  for  his  services,  demanded  only  a  cent  for  every  file  of 
10  men  which  he  could  make  with  a  body  of  100  men.  What 
would  his  demand  amount  to?       Ans.  $  173103094564.40. 


ANALYSIS  BY  POSITION. 

584.  Analysis  by  Position  is  the  process  of  solving 
analytical  questions,  by  assuming  or  supposing  one  or  more 
numbers,  and  reasoning  from  them,  operated  upon  as  if  they 
were  the  number  or  numbers  required  to  be  found. 

585 •    Questions  in  which  the  required  number  is  in  any  way 
increased  or  diminished  in  any  given  ratio,  or  in  which  it  is 
multiplied  or  divided  by  any  number,  may  be  solved  by  means 
of  a  single  assumption. 
35* 


414  ANALYSIS    BY  POSITION. 

586.  Questions  in  which  the  required  numbers,  or  their 
parts,  or  their  multiples,  are  increased  or  diminished  by  some 
given  number  which  is  no  known  part  or  multiple  of  the  re- 
quired number,  or  when  a  power  or  root  of  the  required  num- 
ber is  either  directly  or  indirectly  contained  in  the  result  given 
in  the  question,  may  be  solved  by  two  assumptions. 

Note.  —  When  the  answer  is  obtained  by  means  of  a  single  assumption,  the 
process  is  called  Position;  and  when  obtained  by  means  of  two  assumptions, 
it  is  called  Double  Position. 

Analysis  by  Position  affords  often  a  very  compendious  method  of  working 
questions,  whose  solution  otherwise,  except  by  algebra,  would  be  lengthy  and 
difficult.  It  is  even  found  very  useful  in  shortening  some  of  the  processes  of 
algebra. 

587.  "When  the  answer  may  be  obtained  by  means  of  a 
single  assumption  or  supposition. 

Ex.  1.  A  schoolmaster,  being  asked  how  many  scholars  he 
had,  replied,  that  if  he  had  as  many  more  as  he  now  has,  and 
half  as  many  more,  he  should  have  200.  Of  how  many  schol- 
ars did  his  school  consist  ?  Ans.  80  scholars. 

operation.  By  having  as  many  more,  and 

Assumed  number,      6  0         half  as  many  more,  he  would  have 

As  many  more,  6  0         had  2i times  tlie  <>rIg.mal  number; 

tt  u?  o  ^  therefore,    the    required    number 

Half  as  many  more,  3  0         „,ust  bc  ag  many  £,  200  containg 

\  5  Q  times  2£,  or  80.     The  same  result 

may  be  obtained,  as  in  the  opera- 

150:200::  60:8  0,  Ans.     tion,  thus  :  We  assume  the  number 

of  scholars  to  be  60 ;  if  to  60  as 
many  more,  and  half  as  many  more,  are  added,  the  sum  is  150.  As 
this  result  has  the  same  ratio  to  the  result  in  the  question  as  the  sup* 
posed  number  has  to  the  number  required,  we  find  the  answer  by  a 
proportion. 

Rule.  —  Assume  any  convenient  number,  and  proceed  with  it  ac- 
cording to  the  nature  of  the  question.  Then,  if  the  result  be  either  too 
much  or  too  little,  as  the  result  found  is  to  the  result  given,  so  will  be  the 
number  assumed  to  the  number  required. 

2.  A  person,  after  spending  -J  and  £  of  his  money,  had  $  60 
left ;  what  had  he  at  first?  Ans.  $  144. 

3.  What  number  is  that,  which,  being  increased  by  £,  £,  and 
^  of  itself,  equals  125? 

4.  A's  age  is  double  that  of  B,  and  B's  is  triple  that  of  C, 
and  the  sum  of  all  their  ages  is  140.  What  is  each  person's 
age  ?  Ans.  A's  84,  B's  42,  C's  14  years. 


ANALYSIS    BY  POSITION.  415 

5.  A  person  lent  a  sum  of  money  at  6  per  cent.,  and  at  the 
end  of  10  years  received  the  amount,  $  560.  What  was  the 
sum  lent?   '  Ans.  $350. 

588 1  When  two  or  more  assumptions  or  suppositions  are 
required  in  finding  the  answer. 

Ex.  1.  A  lady  purchased  a  piece  of  silk  at  80  cents  per 
yard,  and  lining  for  it  at  30  cents  per  yard  ;  the  silk  and  lining 
contained  15  yards,  and  the  price  of  the  whole  was  $  7.  How 
many  yards  were  there  of  each  ? 

Ans.  5  yards  of  silk  ;  10  yards  of  lining. 

OPERATION. 

Assume  6  yards  of  silk,  $  4.80     Assume  4  yards  of  silk,  $  3.20 
Lining  would  be  9yd.,       2.70     Lining  would  be  11yd.,      3.30 

Their  sum,  $7.50     Their  sum,  $6.50 

Sum  in  the  question,  7.00     Sum  in  the  question,  7.00 

First  error,  +  $  °-50     Second  error,  —  $  0.50 

.50  -j-  .50  :  6  —  4  : :  .50  :  1 ;  6  yards  —  1  yard  =  5  yards ; 
15  yards  —  5  yards  =  10  yards. 

Since  the  silk  and  lining  contain  15  yards,  cost  $  7.00,  the  average 
price  per  yard  is  46§  cents;  and  80  cents  —  4G§  cents  =  33i  cents; 
46§  cents  —  30  cents  —  16§  cents;  and  the  quantity  of  lining  will 
therefore  be  to  that  of  the  silk  as  33^-  is  to  16|,  or  as  to  2  is  to  1. 
Hence,  if  the  given  number  of  yards,  15,  be  divided  into  3  parts,  two 
of  those  parts,  or  10  yards,  will  be  for  the  lining,  and  the  other  part, 
or  5  yards,  will  be  for  the  silk. 

The  same  result  is  obtained  as  in  the  operation,  thus :  We  assume 
the  quantity  of  silk  to  be  6  yards ;  then  the  lining  would  be  9  yards. 
We  find  the  cost  of  each  of  these  at  the  prices  given,  and,  adding, 
have  as  the  sum  of  their  costs  $  7.50.  This  is  a  result  too  large  by 
$  0.50,  when  compared  with  the  sum  in  question.  By  assuming  the 
quantity  of  the  silk  to  be  4  yards,  and  proceeding  in  a  like  manner, 
we  obtain  for  the  cost  of  the  silk  and  the  lining  $6.50,  a  sum  too 
small  by  $0.50.  The  first  error,  arising  from  a  result  too  large,  is 
marked  with  the  sign  plus  (-)-),  and  the  second  error,  arising  from  a 
result  too  small,  is  marked  with  the  sign  minus  ( — ).  Then,  since 
one  of  the  results  is  too  large,  and  the  other  too  small,  we  then  say, 
as  the  sum  of  the  errors,  or  .50  -)-  .50,  is  to  the  difference  of  the  as? 
sumptions,  or  6  — 4,  so  is  .50  the  less  error,  to  1,  the  correction; 
which,  being  added  to  4  (yards),  the  sum,  5,  expresses  the  number  of 
yards  of  silk;  and  consequently  that  of  the  lining  must  be  10  yards. 

Rule.  —  Assume  two  different  numbers,  perform  on  them  separately 
the  operations  indicated  in  the  question,  and  note  the  errors  of  the  re- 


416  ANALYSIS  BY  POSITION. 

suits.  Then,  as  the  difference  of  the  errors,  if  the  results  be  both  too 
great  or  both  too  small,  or  as  the  sum  of  the  errors,  if  one  result  be  too 
great  and  the  other  too  small,  is  to  the  difference  of  the  assumed  num- 
bers, so  is  either  error  to  the  correction  to  be  applied  to  the  number  which 
produced  that  error. 

Note  1.  —  The  rule  usually  given  fails  in  an  "important  class  of  questions; 
but  the  rule  here  given,  if  not  the  simplest  in  the  resolution  of  some  questions, 
has  the  advantage  of  being  applicable  in  every  case. 

Note  2.  —  In  relation  to  all  questions  which  in  algebra  would  be  resolved 
by  equations  of  the  first  degree,  the  differences  between  the  true  and  the  as- 
sumed numbers  are  proportional  to  the  differences  between  the  result  given  in 
the  question  and  the  results  arising  from  the  assumed  numbers.  But  the  prin- 
ciple does  not  hold  exactly  in  relation  to  other  questions ;  hence,  when  applied 
to  them,  the  above  rule,  or  any  other  of  the  kind  that  can  be  given,  will  only 
produce  approximations  to  the  true  results.  In  which  case  the  assumed  num- 
bers should  be  taken  as  nearly  true  as  possible.  Then,  to  approximate  more 
nearly  to  the  required  number,  assume  for  a  second  operation  the  number 
found  by  the  first,  and  that  one  of  the  first  two  assumptions  which  was  nearer 
the  true  answer,  or  any  other  number  that  may  appear  to  be  still  nearer  to  it. 
In  this  way,  by  repeating  the  operation  as  often  as  may  be  necessary,  the  true 
results  may  be  approximated  to  any  assigned  degree  of  accuracy.  This  pro- 
cess is  sometimes  applied  with  advantage  in  extracting  the  higher  roots,  when 
approximate  results,  differing  but  slightly  from  entire  correctness,  will  answer. 

2.  A  and  B  invested  equal  sums  in  trade ;  A  gained  a  sum 
equal  to  £  ot  his  stock,  and  B  lost  $  225  ;  then  A's  money  was 
double  that  of  B's.     What  did  each  invest  ?  Ans.  $  GOO. 

3.  A  person,  being  asked  the  age  of  each  of  his  sons,  replied, 
that  his  eldest  son  was  4  years  older  than  the  second,  his  sec- 
ond 4  years  older  than  the  third,  his  third  4  years  older  than 
the  fourth,  or  youngest,  and  his  youngest  half  the  age  of  the 
oldest.     What  was  the  age  of  each  of  his  sons  ? 

Ans.  12,  16,  20,  and  24  years. 

4.  A  gentleman  has  two  horses,  and  a  saddle  worth  $  50. 
Now  if  the  saddle  be  put  on  the  first  horse,  it  will  make  his  value 
double  that  of  the  second  horse ;  but  if  it  be  put  on  the  second, 
it  will  make  his  value  triple  that  of  the  first.  What  was  the 
value  of  each  horse  ?        Ans.  The  first,  $  30 ;  second,  $  40. 

5.  A  gentleman  was  asked  the  time  of  day,  and  replied,  that 
§  of  the  time  past  from  noon  was  equal  to  ■£§  of  the  time  to 
midnight.    What  was  the  time  ?         Ans.  12  minutes  past  3. 

6.  A  and  B  have  the  same  income.  A  saves  y1^  of  his, 
but  B,  by  spending  $100  per  annum  more  than  A,  at  the 
end  of  10  years  finds  himself  $  GOO  in  debt.  What  was  their 
income  ?  Ans.  $  480. 


SCALES  OP  NOTATION.  417 

7.  A  gentleman  hired  a  laborer  for  90  days  on  these  condi- 
tions :  that  for  every  day  he  wrought  he  should  receive  60 
cents,  and  for  every  day  he  was  absent  he  should  forfeit  80 
cents.  At  the  expiration  of  the  term  he  received  $  33.  How 
many  days  did  he  work,  and  how  many  days  was  he  idle  ? 

Ans.  He  labored  75  days,  and  was  idle  15  days. 

8.  There  is  a  fish  whose  head  weighs  15  pounds,  his  tail 
weighs  as  much  as  his  head  and  £  of  his  body,  and  his  body 
weighs  as  much  as  his  head  and  tail.  What  is  the  weight  of 
the  fish?  '  Ans.  721b. 

9.  If  12  oxen  eat  3£  acres  of  grass  in  4  weeks,  and  21  oxen 
eat  10  acres  in  9  weeks,  how  many  oxen  would  it  require  to  eat 
24  acres  in  18  weeks,  the  grass  growing  uniformly  ? 

Ans.  36  oxen. 

10.  What  number  exceeds  three  times  its  square  root  by 
11  ?   (Note  2.)  Ans.  26.4201648. 


SCALES    OF   NOTATION. 

589.  The  scale  of  any  system  of  notation  is  the  law  of 
relation  existing  between  its  units  of  different  orders. 

590.  The  radix  of  any  scale  is  the  number  of  units  it  takes 
of  one  order  to  make  a  unit  of  the  next  higher.  Thus,  10  is 
the  radix  of  the  decimal  or  denary  system,  2  of  the  binary,  3  of 
the  ternary,  4  of  the  quaternary,  5  of  the  quinary,  6  of  the 
senary,  7  of  the  septenary,  8  of  the  octary,  9  of  the  nonary, 
11  of  the  undenary  or  undecimal,  12  of  the  duodenary  or  duo- 
decimal,  20  of  the  vigesimal,  30  of  the  trigesimal,  60  of  the  sex- 
agesimal, and  100  of  the  centesimal, 

591 .  In  writing  any  number  in  a  uniform  scale,  as  many 
distinct  characters  or  symbols  are  required  as  there  are  units  in 
the  radix  of  the  given  system.  Thus,  in  the  decimal  or  denary 
scale  10  characters  are  required,  in  the  binary  scale  2  charac- 
ters, in  the  duodenary  or  duodecimal  12  characters,  and  so  on. 
In  the  binary  scale  use  is  made  of  the  characters  1  and  0,  in 


418 


SCALES  OP  NOTATION. 


6)75432 

OPERATION. 

12)75432 

6)12572  0 

12)6286     0 

6)2095  2 

12)523  11,  or  e 

6)349  1 

12)43     7 

6)58  1 

3     7 

6)9  4 

1  3 

the  ternary,  1,  2,  and  0,  &c,  the  cipher  being  one  of  the  char- 
acters in  each  scale.  In  the  duodenary  scale,  eleven  charac- 
ters being  required  beside  the  cipher,  the  first  nine  may  be 
supplied  by  the  nine  digits,  the  tenth  by  t,  the  eleventh  by  e, 
and  the  twelfth  by  0. 

592.  To  change  any  number  expressed  in  the  decimal 
scale  to  any  other  required  scale  of  notation. 

Ex.  1.  Express  the  common  number  75432,  in  the  senary 
and  duodenary  scales.  Ans.  1341120  and  377e0. 

By  dividing  the  given 
number  by  6,  it  is  distribut- 
ed into  12572  classes,  each 
containing  6,  with  0  remain- 
der. By  the  second  division 
by  6,  these  classes  are  dis- 
tributed into  2095  classes, 
each  containing  6  times  6, 
or  the  second  power  of  6, 
with  a  remainder  of  2  of  the 
former  class,  each  containing 
6.  By  the  third  division  the 
classes  last  found  are  distributed  into  349  classes,  each  containing 
6  of  the  latter,  which  were  each  the  second  power  of  6,  and  therefore 
these  are  the  third  power  of  6,  with  a  remainder  1  time  the  second 
power  of  6 ,  In  like  manner,  the  next  quotient  expresses  58  times  the 
fourth  power  of  6,  with  a  remainder  1  time  the  third  power  of  6  ; 
the  next  quotient  expresses  9  times  the  fifth  power  of  6,  with  a  re- 
mainder 4  times  the  fourth  power  of  6 ;  and  the  last  quotient  ex- 
presses 1  time  the  sixth  power  of  6,  with  a  remainder  3  times  the 
fifth  power  of  6.  Hence,  the  given  number  is  found  to  be  equal  to 
1  X  66  +  3  X  65  +  4  X  64  +  1  X  63  +  1  X  62  +  2  X  6  +  0,  or 
according  to  the  senary  system  of  notation  1341120. 

By  proceeding  in  like  manner,  we  find  the  given  number  to  be 
equal  to  3  X  124  +  7  X  123  +  7  X  122  +  11  X  12  +  0,  or,  ac- 
cording to  the  duodenary  scale,  377e0. 

Rule.  — Divide  the  given  number  by  the  radix  of  the  required  scale 
repeatedly ',  till  the  quotient  is  less  than  the  radix ;  then  the  last  quotient, 
icith  the  several  remainders  in  the  retrograde  order  annexed,  placing 
ciphers  where  there  is  no  remainder,  will  be  the  the  given  number  ex- 
pressed in  the  required  scale, 

2.  Change  37  from  the  decimal  to  the  binary  scale. 

Ans.  100101. 

3.  Reduce  1000000  in  the  decimal  scale  to  the  ternary  and 
also  to  the  nonary.  Ans.  1212210202001,  and  1783661. 


SCALES   OP   NOTATION.  419 

4.  How  will   476897  in  the  decimal  scale  be  expressed  in 
the  duodecimal  scale  ?  Ans.  Itee95. 

593 •     To  change  any  number  into  the  decimal  scale,  when 
expressed  in  any  other  scale  of  notation. 

Ex.  1.     Change  377#)  from  the  duodecimal  to  the  decimal 
scale.  Ans.  75432. 


OPERATION. 


We  multiply  the  left-hand  figure  by  the  ra- 
dix, and  add  to  the  product  the  next  figure  ; 
then  we  multiply  this  sum  by  the  radix,  and 
add  to  the  product  the  next  figure,  and  so 
proceed  till  all  the  figures  have  been  em- 
ployed ;  and  we  thus  have,  as  the  values  of  the 
several  figures  collected  into  one  sum,  75432, 
obtained  m  a  manner  similar  to  the  reduction 
of  compound  numbers. 


75432 


Rule.  —  Multiply  the  left-Jiand  figure  of  the  given  number  by  the 
given  radix,  and  to  the  product  add  the  next  figure ;  then  multiply 
this  sum  by  the  radix,  and  add  to  this  product  the  next  figure;  and  so 
proceed  till  all  the  figures  of  the  given  number  have  been  added.  The 
result  will  be  the  given  number  in  the  decimal  scale. 

Note.  —  When  it  is  required  to  change  a  number  from  a  scale  other  than 
decimal  to  another  scale  also  other  than  decimal,  first  change  the  number  as 
given  into  the  decimal  scale,  and  then  the  result  into  the  required  scale. 

2.  Reduce  234  from  the  quinary  to  the  decimal  scale. 

Ans.  69. 

3.  Change  21122  from  the  ternary  to  the  decimal  scale. 

Ans.  206. 

4.  Change  100101  in  the  binary  scale  to  a  number  in  the 
decimal  scale.  Ans.  37. 

5.  Reduce  13579  in  the  duodecimal  scale  to  the  undecimal 
scale.  Ans.  190^3. 

6.  How  will  123454321  in  the  senary  scale  be  expressed  in 
the  duodenary  scale?  Ans.  9873d. 

594  #  To  perform  addition,  subtraction,  multiplication,  di- 
vision, &c.  in  a  scale  of  notation  whose  radix  is  other  than 
10,  we  may 

Proceed  as  in  the  common  scale  of  notation,  except  that  the  radix  of 
the  given  scale  must  be  used  in  the  cases  wherein  the  number  10  would 
be  applied  in  the  decimal  system. 


420  DUODECIMALS. 

Ex.  1.  Required  the  sum  and  difference  of  45324502  and 
25405534  in  the  senary  scale,  or  scale  whose  radix  is  6. 

Ans.  Sum,  115134440;  difference,  15514524. 

2.  Multiply  2483  by  589  in  the  undenary  scale,  or  scale 
whose  radix  is  11.  Ans.  13122^5. 

3.  Divide  1184323  by  589  in  the  duodenary  scale,  whose 
radix  is  12.  Ans.  2483. 

4.  Extract   the   square   root   of    11122441    in   the   senary 
scale.  Ans.  2405. 


DUODECIMALS. 


595.  Duodecimals  are  numbers  expressed  in  a  scale 
whose  radix  is  12,  so  that  12  units  of  each  lower  order  make 
a  unit  of  the  next  higher. 

596.  In  finding  the  contents  of  surfaces  and  solids,  how- 
ever, it  is  customary  to  apply  the  term  duodecimal  to  a  mixture 
of  the  decimal  and  duodecimal  scales.  Thus,  in  admeasure- 
ments in  which  the  foot  is  the  leading  unit,  though  the  differ- 
ent orders  of  units  are  expressed  according  to  the  duodecimal 
scale,  the  number  of  units  in  each  order  is  usually  expressed 
according  to  the  decimal  scale. 

597.  According  to  this  mixed  scale,  the  foot  is  divided 
into  12  equal  parts,  and  each  of  these  parts  into  12  other  equal 
parts,  and  so  on  indefinitely,  giving  y1^,  T¥f,  &c.  In  writing 
these  fractions  without  their  denominations,  to  distinguish  their 
orders,  or  denominations,  accents,  called  indices,  are  written  on 
the  right  of  the  numerators.  Thus,  inches  are  called  primes, 
and  are  marked  ';  the  next  subdivision  is  called  seconds, 
marked  " ;  the  next  is  thirds,  marked  '" ;  and  so  on. 

Note.  —  Numbers  expressed  by  the  mixed  scale  of  feet,  primes,  seconds, 
&c.  may  be  changed  to  the  pure  duodecimal  scale,  and  the  operations  of  addi- 
tion, subtraction,  multiplication,  division,  and  so  on,  then  be  performed  with 
them,  as  in  Art.  594,  observing  to  place  a  point  between  the  unit  and  its  lower 
duodecimal  orders,  and  in  the  result  changing  the  figures  on  the  left  of  the 
point  into  the  decimal  scale,  and  marking  those  on  the  right  as  primes,  sec- 
onds, &c,  according  to  their  places  from  the  order  of  units. 

But  the  operations  of  adding,  subtracting,  &c.  are  usually  performed  by 
other  methods,  such  as  are  given  in  the  articles  that  follow. 


DUODECIMALS.  421 


ADDITION  AND  SUBTRACTION  OF  DUODECIMALS. 

598.     Duodecimals  may  be  added  and  subtracted  in  the 
same  manner  as  compound  numbers. 

Ex.  1.  Add  together  121ft.  3'  9",  105ft  11'  8",  80ft.  0'  6", 
and  15ft.  10'  0"  4"'.  Ans.  323ft.  V  11"  4'". 

2.  From  462ft.  4'  9"  take  307ft.  9'  1". 

3.  What  is  the  value  of  92ft.  0'  6"  —  21ft.  9'  10"  -f-  19ft. 
10'  3"  6"'?  Ans.  90ft.  0'  11"  6'". 


MULTIPLICATION    OF  DUODECIMALS. 

599.  The  index  of  the  unit  of  a  product  of  any  two  duode- 
cimal orders  is  equal  to  the  sum  of  the  indices  of  those  factors. 
That  is,  feet  multiplied  by  a  number  denoting  feet  produces 
feet ;  feet  by  a  number  denoting  primes  ^produces  primes ; 
primes  by  a  number  denoting  primes  produces  seconds,  &c. 

Note.  —  In  multiplication  of  duodecimals,  or  in  other  multiplication,  the 
multiplier  is  always  regarded  as  an  abstract  number,  though  the  notation  of 
feet,  primes,  &c.  is  usually  retained,  in  order  the  better  to  note  the  different 
orders  of  units.  For  the  same  reason,  in  division  of  duodecimals,  the  divisor 
usually  retains  the  notation  of  feet,  primes,  &c. 

600.  To  multiply  one  duodecimal  by  another. 

Ex.  1.  Required  the  number  of  square  feet  in  a  platform  6 
feet  8  inches  long,  and  4  feet  5  inches  wide. 

Ans.  29  sq.  ft.  5'  4". 

We  first  multiply  each  of  the  terms  in  the  multi- 
plicand by  the  5'  in  the  multiplier ;  thus,  8'X  5'=40" 
=  3'  and  4".  Writing  the  4"  under  the  multiplier, 
we  reserve  the  3'  to  add  to  the  next  product. 
Then  6ft.  X  5'  =  30' ;  and  30'  -|-  3'  =  33'  =  2ft. 
and  9',  which  we  write  in  their  order  beneath  the 
multiplier.  We  next  multiply  by  the  4ft.,  thus: 
29sq.ft.  5'  4"  8'  X  4  feet  =  32'  =  2ft.  and  8'.  We  write  the  8' 
under  the  primes  in  the  other  partial  product,  and 
reserve  the  2ft.  to  add  to  the  next  product ;  and  6ft.  X  4$.  =  24ft. ; 
24ft.  -\-  2ft.  =  26ft.,  which  we  write  under  the  feet  in  the  other  par- 
tial product.  The  two  being  added  together,  we  have  29  sq.  ft.  5'  4"; 
or  (the  primes  and  seconds  being  changed  to  a  fraction  of  a  foot), 
29/^sq.  ft. 

N     36 


FIRST 

OPERATION. 

6ft. 

8' 

4ft. 

5' 

2 

9/4// 

26 

8' 

SECOND    OPERATION, 

6-8 

4-5 

294 

228 

422  DUODECIMALS. 

In  the  second  operation  the  work 
is  performed  as  in  pure  duodecimals 
(Art.  594).  The  point  (•)  separates 
lower  duodecimal  orders  from  those  be- 
ginning with  feet.  As  to  the  number 
of  feet  in  the  multiplicand  and  multi- 
plier, no  change  is  required  in  that 
2  5*5  4  =  29sq.  ft.  o'  &.",     part  0f  either  of  the  given  numbers  in 

expressing  them  according  to  the  duo- 
decimal scale.  In  performing  the  multiplication,  we  make  the  several 
reductions  required  according  to  the  radix  1 2 ;  and  have,  after  point- 
ing off,  25*54  square  feet,  expressed  according  to  the  duodecimal 
scale.  The  units,  or  feet,  at  the  left  of  the  point,  are  readily- 
changed  to  the  decimal  scale  by  multiplying  the  left-hand  figure,  2, 
by  12,  the  number  of  units  in  the  radix,  and  adding  the  right-hand 
figure,  5,  and  giving  the  figures  to  the  right  of  the  point  their  proper 
notation,  we  have  then  29sq.  ft.  5'  4"  for  the  answer,  as  before. 

Rule.  —  Write  the  multiplier  under  the  multiplicand,  so  that  units 
of  the  same  orders  shall  stand  in  the  same  column. 

Beginning  at  the  right,  multiply  each  term  in  the  midtiplicand  by  each 
term  of  the  multiplier,  and  write  the  first  term  of  each  partial  product 
under  its  multiplier,  observing  to  carry  a  unit  for  every  twelve  from 
each  lower  order  to  the  next  higher. 

The  sum  of  the  partial  product  will  be  the  product  required. 

2.  How  many  square  feet  in  a  floor  48  feet  6  inches  long, 
and  24  feet  3  inches  broad  ?  Ans.  1176  sq.  ft.  1'  G". 

3.  The  length  of  a  room  being  20  feet,  its  breadth  14  feet 
6  inches,  and  height  10  feet  4  inches,  how  many  square  yards 
of  painting  are  in  it,  deducting  a  fireplace  of  4  feet  by  4  feet  4 
inches,  and  2  windows,  each  6  feet  by  3  feet  2  inches  ? 

Ans.  73/T  square  yards. 

4.  Required  the  solid  contents  of  a  wall  53  feet  6  inches 
long,  10  feet  3  inches  high,  and  2  feet  thick. 

5.  There  is  a  house  with  four  tiers  of  windows,  and  four 
windows  in  a  tier ;  the  height  of  the  first  is  6  feet  8  inches ; 
the  second,  5  feet  9  inches ;  the  third,  4  feet  6  inches  ;  the 
fourth,  3  feet  10  inches ;  and  the  breadth  is  3  feet  5  inches ; 
how  many  square  feet  do  they  contain  in  the  whole  ? 

Ans.  283sq.  ft.  Tin. 

6.  How  many  cords  in  a  pile  of  wood  97  feet  9  inches  long, 
4  feet  wide,  and  3  feet  6  inches  high  ?         Ans.  10^£  cords. 

7.  Required  the  number  of  cords  of  wood  in  a  pile  100  feet 
long,  4  feet  wide,  and  6  feet  11  inches  high.  Ans.  21 1|. 


DUODECIMALS.  423 


DIVISION    OF    DUODECIMALS. 

601 .     To  divide  one  duodecimal  by  another. 

Ex.  1.  A  board  in  the  form  of  a  rectangle,  whose  area  is 
27  sq.  ft.  8'  6",  is  1ft.  7in.  wide ;  what  is  its  length  ? 

Ans.  17ft.  Gin. 
first  operation.  We  find  how  many  times  27 

lft.  7')  27sq.ft.  8'  6"  ( 17ft.  6'      square  feet  contains  the  divisor, 
26  11  and  obtain  17  feet  for  the  quo- 

tient,  we  multiply  the  entire  di- 

9    6  visor  by  the  17ft.,  and  subtract 

9    6  the  product,  26ft.  11',  from  the 

corresponding  portion  of  the  div- 
idend, and  obtain  9',  to  which  remainder  we  bring  down  the  6", 
and,  dividing,  obtain  6'  for  the  quotient.  Multiplying  the  entire 
divisor  by  6",  we  obtain  Oft.  9'  6",  which,  subtracted  as  before,  leaves 
no  remainder.     Therefore  1 7  feet  6  inches  is  the  length  required. 

second  operation.  In  the  second  operation  we  re- 

1-7  )  23-86  (  15-6  as  17ft.  6'      duce  the  feet  of  the  given  multi- 

yj  plicand  and  multiplier  to  the  duo- 

■ decimal    scale,    and    thus    obtain 

88  23-86  and  1-7.     We  then  conduct 

7  e  the  division  with  reference  to  the 

^TT  radix  12,  as  is  ordinarily  done  with 

^  respect  to  10  (Art.  594  ).    The  re- 

^k  suit  obtained  is  15-6  in  the  duo- 

decimal, which,  on  changing  the 
figures  to  the  left  of  the  point  to  the  decimal  scale,  and  giving  the 
proper  notation  to  the  figure  on  the  right  of  the  point,  becomes  trans- 
formed to  17ft.  6'  =  17ft.  6in.,  the  answer,  as  before. 

Rule.  —  Find  how  many  times  the  highest  term  of  the  dividend  will 
contain  the  divisor.  By  this  quotient  multiply  the  entire  divisor,  and 
subtract  the  product  from  the  corresponding  terms  of  the  dividend.  To 
the  remainder  annex  the  next  denomination  of  the  dividend,  and  divide 
as  before,  and  so  continue  till  the  division  is  complete. 

2.  It  required  834  sq.  ft.  3'  of  board  to  cover  the  side  of  a 
certain  building.  The  height  was  17ft.  9in. ;  what  was  the 
length  of  the  side  ?  Ans.  47  feet. 

3.  How  many  feet  wide  is  a  plank  of  uniform  width,  whose 
length  is  18ft  9in.,  thickness  3  inches,  and  solid  contents  84ft. 
4'  6"? 

4.  An  alley  has  an  area  of  792ft.  6'  9"  2//;.  Its  width  is 
12ft.  7'  8".     Eequired  its  length.  Ans.  62ft.  8'  6". 


'424  MISCELLANEOUS   EXAMPLES, 


MISCELLANEOUS   EXAMPLES. 

1.  A  merchant  engages  a  clerk  at  the  rate  of  $  20  for  the 
first  month,  $  25  for  the  second,  $  30  for  the  third,  &c,  thus  in- 
creasing his  salary  by  $  5  per  month.  How  long  must  the 
clerk  retain  his  situation,  so  as  to  receive  on  the  whole  as  much 
as  he  would  have  received  had  his  salary  been  fixed  at  $  52.50 
per  month  ?  Ans.  14  months. 

2.  A  mason  has  plastered  3  rooms ;  the  ceiling  of  each  is  20 
feet  by  16  feet  6  inches,  the  walls  of  each  are  9  feet  6  inches 
high,  and  90  yards  are  to  be  deducted  for  doors,  windows,  &c. 
For  how  many  yards  must  he  be  paid  ?    Ans.  251yd.  1ft.  Gin. 

3.  A  man  of  wealth,  dying,  left  his  property  to  his  ten  sons, 
and  the  executor  of  his  will,  as  follows :  to  his  executor, 
$  1024  ;  to  his  youngest  son,  as  much  and  half  as  much  more  ; 
and  increasing  the  share  of  each  next  elder  in  the  ratio  of  1J-. 
What  was  the  share  of  the  eldest? 

4.  A  butcher,  wishing  to  buy  some  sheep,  asked  the  owner 
how  much  he  must  give  him  for  20 ;  on  hearing  his  price,  he 
said  it  was  too  much ;  the  owner  replied,  that  he  should  have 
10,  provided  he  would  give  him  a  cent  for  each  different  choice 
of  10  in  20,  to  which  he  agreed.  How  much  did  he  pay  for  the 
10  sheep,  according  to  the  bargain  ?  Ans.  $  1847.56. 

5.  If  340  square  feet  of  carpeting  are  required  to  cover  the 
floor  of  a  room,  how  many  yards  will  be  required,  provided  the 
width  of  the  carpeting  is  3  feet  9  inches  ?        Ans.  30yd.  8in. 

6.  If  a  clergyman's  salary  of  $  700  per  annum  is  6  years  in 
arrears,  how  much  is  due  him,  allowing  compound  interest  at  6 
per  cent.  ?  Ans.  $  4882.72. 

7.  Suppose  a  clock  to  have  an  hour-hand,  a  minute-hand, 
and  a  second-hand,  all  turning  on  the  same  centre.  At  12 
o'clock  all  the  hands  are  together  and  point  at  12. 

(1.)  How  long  will  it  be  before  the  second-hand  will  be  be- 
tween the  other  two  hands,  and  at  equal  distances  from 
each?  Ajis.  60t7^°7  seconds. 

(2.)  Also  before  the  minute-hand  will  be  equally  distant  be- 
tween the  other  two  hands?  Ans.  61fff  seconds. 

(3.)  Also  before  the  hour-hand  will  be  equally  distant  be- 
tween the  other  two  hands  ?  Ans.  59^-f  seconds. 


MENSURATION.  425 

MENSURATION. 

DEFINITIONS. 

602. '    A  point  is  that  which  has  neither  length,  breadth,  nor  thick- 
ness, but  position  only.  » 

603.     A  line  is  length,  without  breadth  or  thickness. 
A  straight  line  is  one  which  has  the  same  direction 
in  its  whole  extent ;  as  the  line  A  B. 

A  curved  line  is  one  which  continually  changes  its 


direction ;  as  the  line  C  D.  C^  \d 

604.  An  angle  is  the  inclination  or  opening  of  two  lines, 
which  meet  in  a  point. 

A 

A  right  angle  is  an  angle  formed  by  a  straight  line  and  a 
perpendicular  to  it ;  as  the  angle  ABC. 

An  acute  angle  is  one  less  than  a  right  angle  ;  as 

the  angle  E  B  C. 

B 

An  obtuse  angle  is  one  greater  than  a  right  angle  ;    ^ 
as  the  angle  F  B  C. 

B 

605.  A  surface  is  that  which  has  length  and  breadth,  without 
thickness. 

A  plane  surface,  or  simply  a  plane,  is  that  in  which,  if  any  two 
points  whatever  be  taken,  the  straight  line  that  joins  them  will  lie 
wholly  in  it. 

Every  surface,  which  is  not  a  plane,  or  composed  of  planes,  is  a 
curved  surface. 

606.  The  area  of  a  figure  is  its  quantity  of  surface ;  and  is  es- 
timated in  the  square  of  some  unit  of  measure,  as  a  square  inch,  a 
square  foot,  &c. 

607.  A  solid,  or  body,  is  that  which  has  length,  breadth,  and 
thickness. 

608.  The  solidity,  or  volume  of  a  solid,  is  estimated  in  the  cube  of 
some  unit  of  measure ;  as  a  cubic  inch,  a  cubic  foot,  &c. 

609.  Mensuration  is  the  process  of  determining  the  areas  of 
surfaces,  and  the  solidity  or  volume  of  solids. 

MENSURATION   OF   SURFACES. 

610.  A  plane  figure  is  an  enclosed  plane  surface  ;  if  bounded  by 
straight  lines  only,  it  is  called  a  rectilineal  figure,  or  polygon.  The 
perimeter  of  a  figure  is  its  boundary,  or  contour. 

36* 


0 


426 


MENSURATION. 


61 1  •    Three-sided  polygons  are  called  triangles ;  those  of  four  sides, 
quadrilaterals  ;  those  of  five  sides,  pentagons,  and  so  on. 

Triangles. 

A 

612.     An  equilateral  triangle  is  one  whose  sides  are 
all  equal ;  as  CAD. 

Notk.  — The  line  AB,  drawn  from  the  angle  A  perpendic- 
ular to  the  base  C  I),  is  the  altitude  of  the  triangle  GAD.  q 


An  isosceles  triangle  is  one  which  has  two  of  its  sides 
equal ;  as  E  F  G. 


A  scalene  triangle  is  one  which  has  its  three 
sides  unequal ;  as  II I  J. 


ii 


A  riglit-angled  triangle  is  one  which  has  a  right 
angle ;  as  KLM. 


613.     To  find  the  area  of  a  triangle. 

j\hdtiphj  the  base  by  half  the  altitude,  and  the  product  will  be  the 
area.     Or, 

Add  the  three  sides  together,  take  half  that  sum,  and  from  this  sub- 
tract each  side  separately ;  then  multiply  the  half  of  the  sum  and  these 
remainders  together,  and  the  square  root  of  this  product  will  be  the  area. 

Ex.  1.  What  are  the  contents  of  a  triangle  whose  perpendicular 
height  is  12  feet,  and  whose  base  is  18  feet  ?  Ans.  108  feet. 

2.  There  is  a  triangle,  the  longest  side  of  which  is  15.6  feet,  the 
shortest  side  9.2  feet,  and  the  other  side  10.4  feet.  What  are  the 
contents  ?  Ans.  46.139-f-  feet 

3.  The  triangular  gable  of  a  certain  building  has  a  base  of  40  feet 
and  an  altitude  of  15  feet ;  how  many  square  feet  of  boards  will  cover 
it  ?  Ans.  300  sq.  ft. 

4.  The  perimeter  of  a  certain  field  in  the  form  of  an  equilateral 
triangle  is  336  rods  ;  what  is  the  area  of  the  field  ? 

Ans.  33  acres  152  sq.  rd. 

Quadrilaterals. 
61 4#     A  parallelogram  is  any  quadrilateral  whose  opposite  sides 

d  , ,  c 


are  parallel. 


615.     A  rectangle  is  any  right-angled  parallel- 
ogram ;  as  ABCD. 


MENSURATION.  427 

H  a 


E 


616.     A  square  is  a  parallelogram  whose  sides  are 
equal,  and  whose  angles  are  right  angles ;  asEFGH. 


61 7#  A  rhombus  is  a  parallelogram  whose  sides 
are  equal,  and  whose  angles  are  not  right  angles ;  as 
IJKL. 


618.      A  rhomboid  is  a  parallelogram  whose  an-        J\  / 

gles  are  not  right  angles ;  as  M  N  O  P.  /  \  / 

M     Q        N 
Note.  —  The  altitude  of  a  parallelogram  is  the  perpendicular  distance  be- 
tween any  two  of  its  parallel  sides  taken  as  bases,  as  the  line  P  Q,  drawn 
between  two  sides  of  the  rhomboid  M  N  0  P,  and  perpendicular  to  the  sides 
MN  and  OP. 

U  T 


K 


Z 


619.    A  trapezoid  is  a  quadrilateral  which  has  only 
two  of  its  sides  parallel ;  as  KSTU. 


620.  A  trapezium  is  a  quadrilateral  which  has 
no  two  sides  parallel ;  asWXYZ. 

w 

Note.  —  A  diagonal  of  a  quadrilateral,  or  of  any  polygon  of  more  than  four 
sides,  is  a  straight  line  which  joins  the  vertices  of  two  opposite  angles,  or  of  two 
angles  not  adjacent;  as  the  line  XZ  joining  vertices  of  opposite  angles  of  the 
trapezium  \Y  X  Y  Z. 

621.  To  find  the  area  of  a  parallelogram. 

Multiply  the  base  by  the  altitude,  and  the  product  will  be  the  area, 

Ex.  1.  What  are  the  contents  of  a  board  15  feet  long  and  2  feet 
wide  ?  Ans.  30  feet. 

2.  A  rectangular  state  is  128  miles  long  and  48  miles  wide.  ^  How 
many  square  miles  does  it  contain  ?  Ans.  6144  miles. 

3.  The  base  of  a  rhomboid  being  12  feet,  and  its  height  8  feet,  re- 
quired the  area.  Ans.  96  feet. 

4.  Required  the  area  of  a  rhombus  of  which  one  of  the  equal  sides 
is  358  feet,  and  the  perpendicular  distance  between  it  and  the  oppo- 
site side  is  194  feet.  Ans.  69452  sq.  ft. 

5.  The  largest  of  the  Egyptian  pyramids  is  square  at  its  base, 
and  measures  693  feet  on  a  side.  How  much  ground  does  it 
cover  ?  Ans.  1 1  acres  4  poles. 

6.  What  is  the  difference  between  the  area  of  a  floor  40  feet 
square,  and  that  of  two  others,  each  20  feet  square  ?     Ans.  800  feet. 

7.  There  is  a  square  whose  area  is  3600  yards ;  what  are  the  sides 


428  MENSURATION. 

of  the  square,  and  the  breadth  of  a  walk  along  each  side  and  each 
end  of  the  square,  which  shall  take  up  just  one  half  of  the  square  ? 

A        (  42.42-4-  yards,  side  of  the  square. 

Ans*  j    8.78+  yards,  breadth  of  the  walk. 

622 •     To  find  the  area  of  a  trapezoid. 

Multiply  half  of  the  sum  of  the  parallel  sides  by  the  altitude,  and  the 
product  is  the  area. 

Ex.  1.  If  the  parallel  sides  of  a  trapezoid  are  75  and  33  feet,  and 
the  perpendicular  breadth  20  feet,  what  is  the  area  ? 

Ans.  1080  sq.  ft. 

2.  Required  the  area  of  a  meadow  in  the  form  of  a  trapezoid, 
whose  parallel  sides  are  786  and  473  links,  and  whose  altitude  is  986 
links.  Ans.  6  acres  33  rods  3  yards. 

623*     To  find  the  area  of  a  trapezium. 

Divide  the  trapezium  into  two  triangles  hy  a  diagonal,  and  then  find 
the  areas  of  these  triangles ;  their  sum  will  be  the  area  of  the  trape- 
zium. 

Ex.  1.  Required  the  area  of  a  garden  in  the  form  of  a  trapezium, 
of  which  the  four  sides  are  328,  456,  572,  and  298  feet,  and  the  diag- 
onal, drawn  from  the  angle  between  the  first  and  second  sides,  598 
feet.  Ans.  3  acres  1  rood  31  rods  29  yards  3.85  feet. 

2.  Given  one  of  the  diagonals  of  a  field,  in  the  form  of  a  trapezium, 
equal  1 7  chains  56  links,  to  compute  the  area,  the  perpendiculars  to 
that  diagonal  from  the  opposite  angles  being  8  chains  82  links,  and  7 
chains  73  links.  Ans.  14  acres  2  roods  5  rods. 

Pentagons,  Hexagons,  &c. 

624.  A  pentagon  is  a  polygon  of  five  sides  ;  a  hexagon,  one  of  six 
sides ;  a  heptagon,  one  of  seven  sides ;  an  octagon,  one  of  eight  sides ; 
a  nonagon,  one  of  nine  sides;  and  so  on  for  a  decagon,  undecagon, 
dodecagon,  &c.  D 

625.  A  regular  polygon  is  one  whose  sides  and 
angles  are  equal ;  as  the  pentagon  ABODE. 

626.  To  find  the  area  of  a  regular  polygon. 

Multiply  the  perimeter  by  half  the  perpendicular  let  fall  from  the 
centre  upon  one  of  the  sides.     Or, 

Multiply  the  square  of  one  of  the  sides  by  the  number  against  the  poly- 
yon  in  the  following 

Table. 


Pentagon, 

1.720477 

Nonagon, 

6.181824 

Hexagon, 

2.598076 

Decagon, 

7.694209 

Heptagon, 

3.633913 

Undecagon, 

9.365641 

Octagon, 

4.828427 

Dodecagon, 

11.196152 

MENSURATION. 


429 


Ex.  1.  What  is  the  area  of  a  regular  pentagon,  of  which  the  side  is 
250  feet,  and  the  perpendicular  from  the  centre  to  one  side  172.05 
feet  ?  Ans.  107531.25  sq.  ft. 

2.  What  is  the  area  of  a  regular  hexagon  whose  side  is  356  yards, 
and  whose  perpendicular  is  308.305  yards  ?  Ans.  329269.74yd. 

3.  The  side  of  a  regular  octagonal  enclosure  is  60  yards ;  how 
many  acres  are  included  ?     Ans.  3  acres  2  roods  14  rods  19  yards. 

4.  The  side  of  a  field,  whose  shape  is  that  of  a  regular  decagon,  is 
243  feet ;  what  is  its  area  ? 

Ans.  10  acres  1  rood  28  rods  24  yards  6.347  feet. 


Circles. 

627  •  A  circle  is  a  plane  figure  bounded  by 
a  line,  every  part  of  which  is  equally  distant  from 
a  point  within  called  the  centre  ;  as  AEFGBD. 


The  circumference  or  periphery  of  a  circle  is  the  line  that 
bounds  it. 

A  radius  is  a  line  drawn  from  the  centre  to  the  circumference  ;  as 
C  A,  or  C  D. 

A  diameter  is  a  line  which  passes  through  the  centre,  and  is  termi- 
nated by  the  circumference  ;  as  A  B. 

An  arc  is  any  portion  of  the  circumference  ;  as  A  D,  A  E,  or  E  G  F. 

The  chord  of  an  arc  is  the  straight  line  joining  its  extremities ;  as 
E  F,  which  is  the  chord  of  the  arc  E  G  F. 

628 1  The  segment  of  a  circle  is  the  portion  included  by  an  arc 
and  its  chord;  as  the  space  included  by  the  arc  EGF  and  the 
chord  E  F. 

629.  The  sector  of  a  circle  is  the  portion  included  by  two  radii 
and  the  intercepted  arc ;  as  the  space  ACDA. 

630  •  A  zone  is  the  space  between  two  parallel  chords  of  a 
circle  ;  as  the  space  A  E  F  B  A. 

O 

631.  A  lune,  or  crescent,  is  the  space 
included  between  the  intersecting  arcs  of 
two  eccentric  circles ;  as  A  C  B  E  A. 


632.     A  circular  ring  is  the  space  included 
between   the  circumference   of  two  concentric  A 
circles  ;  as  the  space  between  the  rings  A  B  and 
CD. 


430  MENSURATION. 

633  •     To  find  the  circumference  of  a  circle,  the  diameter  being 
given. 

Multiply  the  diameter  by  3.141592. 

Ex.  1.     If  the  diameter  of  a  circle  is  144  feet,  what  is  the  circum- 
ference ?  Ans.  452.389248  feet. 

2.  If  the  diameter  of  the  earth  is  7964  miles,  what  is  its  circum- 
ference? Ans.  25019.638688-j-  miles. 

3.  Required  the  circumference  of  a  circle,  whose  radius  is  512 
feet.  Ans.  4  furlongs  34  rods  5  yards  1  foot. 

634.     To  find  the  diameter  of  a  circle,  the  circumference  being 
given. 

Multiply  the  circumference  by  .318309. 

Ex.  1.  Required  the  diameter  of  a  circle,  whose  circumference 
is  1043  feet.  Ans.  331.997-J-  feet. 

2.  If  the  circumference  of  a  circle  is  25000  miles,  what  is  its  diam- 
eter? Ans.  7957.74-}- miles. 

3.  If  the  circumference  of  a  round  stick  of  timber  is  50  inches, 
what  is  its  diameter  ?  Ans.  15.91549-f-  inches. 

635*  To  find  the  area  of  a  circle,  the  diameter,  or  the  circum- 
ference, or  both,  being  given. 

Multiply  the  square  of  the  diameter  by  .785398.     Or, 
Multiply  the  square  of  the  circumference  by  .079577.     Or, 
Multiply  half  the  diameter  by  half  the  circumference. 

Ex.  1.    If  the  diameter  of  a  circle  is  761  feet,  what  is  the  area  ? 

Ans.  454840.475158  feet 

2.  There  is  a  circular  island,  three  miles  in  diameter ;  how  many- 
acres  does  it  contain  ?  Ans.  4523. 89-f-  acres. 

3.  Required  the  area  of  a  circle,  of  which  the  circumference  is  1284 
yards.  Ans.  27  acres  17  rods  0.8-j-  yards. 

4.  Required  the  area  of  a  circle,  of  which  the  diameter  is  169, 
and  the  circumference  532  inches.   Ans.  17  yards  3  feet  13  inches. 

636.     To  find  the  area  of  a  sector  of  a  circle. 
Multiply  the  length  of  the  arc  by  half  the  radius  of  the  circle.     Or, 
As  360°  are  to  the  degrees  in  the  arc  of  the  sector,  so  is  the  area  of  the 
circle  to  the  area  of  the  sector. 

Ex.  1 .  Required  the  area  of  a  sector,  of  which  the  arc  is  79  and 
the  radius  of  the  circle  47  inches.  Ans.  1856.5  inches. 

2.  Required  the  area  of  a  sector,  of  which  the  arc  is  26°,  and  the 
radius  of  the  circle  25  feet.  Ans.  141.8  square  feet. 

637  •     To  find  the  area  of  the  segment  of  a  circle. 

Find  the  area  of  the  sector  which  has  the  same  arc  with  the  segment ; 
and  also  the  area  of  the  triangle  formed  by  the  chord  and  the  radii 
drawn  to  its  extremities.  The  difference  of  these  areas,  ichen  the  seg- 
ment is  less,  and  their  sum,  when  the  segment  is  greater,  than  the  semi- 
circle, will  be  the  area  of  the  segment.     Or, 


I 

MENSURATION.  431 

To  two  thirds  of  the  'product  of  the  height  of  the  segment  by  the  chord 
add  the  cube  of  the  height,  divided  by  twice  the  chord, 

Ex.  1.  Required  the  area  of  the  segment 
ABC  A,  of  which  the  arc  ABC  is  49.25°,  the 
chord  AC  10  feet,  and  the  radii  E  A,  E  B,  and 
E  C,  each  12  feet.  Ans.  7.35  sq.  ft. 


2.  Required  the  area  of  a  segment  whose  height  is  15  rods  and 
whose  chord  is  24  rods.  Ans.  1  acre  3  roods  30  rods  9.4  yards. 

638.     To  find  the  area  of  a  zone  of  a  circle. 

From  the  area  of  the  whole  circle  subtract  the  areas  of  the  segments 
on  the  sides  of  the  zone, 

Ex.  1.  Required  the  area  of  a  zone  whose  parallel  sides  are  23.25 
and  20.8  feet,  in  a  circle  whose  radius  is  12  feet.       Ans.  206  sq.  ft. 

2.  Required  the  area  of  a  zone  included  between  two  chords  of  1 6 
feet  each,  the  diameter  of  the  circle  being  20  feet.    Ans.  224.7  sq.  ft. 

639 #     To  find  the  area  of  a  lune  or  crescent. 

Find  the  difference  of  the  areas  of  the  two  segments  formed  by  the 
arcs  of  the  lune  and  its  chord. 

Ex.  1.  If  the  chord  of  two  intersecting  arcs  is  72  feet,  and  the 
height  of  one  of  the  segments  is  30,  and  of  the  other  20  feet,  what  is 
the  area  of  the  crescent  ?  Ans.  612  sq.  ft. 

640 1     To  find  the  area  of  a  circular  ring. 

Multiply  the  sum  of  the  diameters  of  the  two  circles  by  the  difference 
of  the  diameters,  and  that  product  by  .78.54. 

Ex.  1.  What  is  the  area  of  the  ring  formed  by  two  circles  whose 
diameters  are  10  and  20  yards?  Ans.  235.62  sq.  yd. 

2.  In  the  centre  of  a  circular  pond  there  is  an  island  128  yards  in 
diameter ;  what  is  the  area  of  the  pond,  provided  the  exact  distance 
from  any  part  of  the  outer  side  of  the  pond  to  the  centre  of  the  island 
is  78^  yards  ?  Ans.  1  acre  1  rood  14  rods  17  yards  7.4  feet. 

641.  To  find  the  side  of  a  square  that  shall  equal  the  area  of  a 
circle  of  a  given  diameter  or  circumference. 

Multiply  the  diameter  of  the  circle  by  .886227.     Or, 

Multiply  the  circumference  of  the  circle  by  .282094. 

Ex.  1.  I  have  a  round  field,  50  rods  in  diameter ;  what  is  the  side 
of  a  square  field  that  shall  contain  the  same  area  ? 

Ans.  44.31135-f-rods. 

2.  I  have  a  circular  field  360  rods  in  circumference ;  what  must  be 
the  side  of  a  square  field  that  shall  contain  the  same  area  V 

Ans.  101.55-j-  rods. 

3.  John  Smith  had  a  farm  which  was  10,000  roils  in  circumference, 
which  he  sold  at  $  71.75  per  acre.     He  purchased  another  farm  con- 


432  MENSURATION. 

taining  the  same  quantity  of  land  in  the  form  of  a  square ;  required  the 
length  of  one  of  its  sides.  Ans.  2820.94-j-  rods. 

642 1  To  find  the  diameter  of  a  circle  that  shall  contain  the  area 
of  a  given  square. 

Multiply  the  side  of  the  given  square  by  1.12838. 

Ex.  1.  The  side  of  a  square  is  44.31135  rods  ;  required  the  diam- 
eter of  a  circular  field  containing  the  same  area. 

643.  To  find  the  side  of  the. largest  equilateral  triangle  that  can 
be  inscribed  in  a  circle  of  a  given  diameter  or  circumference. 

Multiply  the  given  diameter  by  .866025.     Or, 
Multiply  the  given  circumference  by  .275664. 

Ex.  1.  There  is  a  certain  piece  of  round  timber  30  inches  in  di- 
ameter ;  required  the  side  of  an  equilateral  triangular  beam  that  may 
be  hewn  from  it.  Ans.  25.98-)-  inches. 

2.  How  large  an  equilateral  triangle  may  be  inscribed  in  a  circle 
whose  circumference  is  5000  feet  ?  Ans.  1378.323  feet 

3.  Required  the  side  of  an  equilateral  triangular  beam,  that  may  be 
hewn  from  a  round  piece  of  timber  80  inches  in  circumference. 

Ans.  22.05-f-  inches. 

644  •  To  find  the  side  of  the  largest  square  that  can  be  inscribed 
in  a  circle  of  a  given  diameter  or  circumference. 

Multiply  the  given  diameter  by  .707106.     Or, 
Multiply  the  given  circumference  by  .225079. 

Note.  —  To  find  the  circumference  of  a  circle  required  to  exactly  admit  a 
square  of  a  given  side,  divide  the  given  side  by  .225079. 

Ex.  1.  I  have  a  piece  of  timber  30  inches  in  diameter ;  how  large 
a  square  stick  can  be  hewn  from  it?  Ans.  21.214-  in.  square. 

2.  Required  the  side  of  a  square  that  may  be  inscribed  in  a  circle 
80  feet  in  diameter.  Ans.  56.56848-f-  fret. 

3.  I  have  a  circular  field  whose  circumference  is  5000  rods ;  what 
is  the  side  of  the  largest  square  field  that  can  be  made  in  it  ? 

Ans.  1125.395+ rods. 

4.  How  large  a  square  stick  may  be  hewn  from  a  piece  of  round 
timber  100  inches  in  circumference  ?        Ans.  22.5-f-  inches  square. 

5.  What  must  be  the  circumference  of  a  tree  that,  when  hewn,  shall 
be  18  inches  square?  Ans.  79.9 7— [—  inches. 

6.  I  have  a  garden  which  is  20  rods  square ;  required,  in  feet,  the 
circumference  of  a  circle  that  will  enclose  this  garden. 

Ans.  1466.15-[-  feet. 

645.  To  find  the  diameter  of  the  three  largest  equal  circles  that 
can  be  inscribed  in  a  circle  of  a  given  diameter. 

Divide  the  given  diameter  by  2.155. 

Ex.  1.  Required  the  diameter  of  each  of  the  largest  three  circles 
that  can  be  inscribed  in  a  circle  86.2  inches  in  diameter. 

Ans.  40  inches. 


MENSURATION. 


433 


Ellipse. 

646.  An  ellipse  is  a  plane  figure  bounded 
by  a  curve,  from  any  point  of  which  the  sum 
of  the  distances  to  two  fixed  points  is  equal  k[  G 
to  a  given  distance.  The  two  fixed  points 
are  called  the  foci,  as  GH  in  the  ellipse 
AECBDFA.  The  major  or  transverse  axis 
of  an  ellipse  is  its  longest  diameter,  as  A  B.  The 
minor  or  conjugate  axis  of  an  ellipse  is  its  shortest  diameter,  as  C  D.  The 
segment  of  an  ellipse  is  a  portion  cut  off  from  the  ellipse,  as  F  A  E  F. 

647«     To  find  the  area  of  an  ellipse,  the   two  diameters  being 
given. 

Multiply  the  two  diameters  together,  and  that  product  by  .785398. 

Ex.  1.     What  is  the  area  of  an  ellipse,  whose  two  diameters  are  24 
and  18  inches  ?  Ans.  339.2928  inches. 

2.  What  is  the  area  of  an  elliptical  pond,  whose  longest  diameter  is 
33  feet  5  inches,  and  whose  shortest  diameter  20  feet  3  inches  ? 

Ans.  59  sq.  yd.  67  sq.  in. 


MENSURATION    OF    SOLIDS. 
Prisms  and  Cylinders. 

648*  A  prism  is  a  figure  whose  ends  or  bases  are  any  plane 
figures  which  are  equal  and  similar,  and  parallel  to  each  other,  and 
whose  sides  are  parallelograms.         /  A 

A  triangular  prism  is  one  whose  base  is  a  triangle  ;  as  the 
figure  A  B. 

A  square  prism  is  one  whose  base  is  a  square  ;  a  pentagonal 
prism,  one  whose  base  is  a  pentagon  ;  and  so  on,  according  to 
the  figure  of  the  ends  or  bases. 

A  parallelopiped  is  a  prism  whose  ends  or  bases,  as  well  as 
its  sides,  are  parallelograms. 

649.  A  cylinder  is  a  round  body  of  uniform  diameter, 
and  which  has  circular  bases  parallel  to  each  other ;  as  the 
figure  C  D. 

The  perimeter  of  a  prism  or  cylinder  is  the  line  that  bounds 
its  end  or  base ;  and  the  altitude  or  height  is  the  distance 
between  the  ends  or  bases. 

The  convex  surface  of  a  prism  or  cylinder  is  the  entire 
surface,  exclusive  of  the  two  ends  or  bases. 

650.  To  find  the  surface  of  a  prism,  or  of  a  cylinder. 

Multiply  the  perimeter  of  the  given  prism  or  cylinder  by  the  height, 
and  to  the  product  add  the  area  of  the  two  ends. 

Ex.  1.  Required  the  surface  of  a  triangular  prism,  of  which  the  dis- 
n     37 


434  MENSURATION. 

tance  between  the  ends  is  13  feet,  and  the  sides  of  the  base  23.34  and 
19  inches.  Ans.  85.22-}-  square  feet. 

2.  Required  the  surface  of  a  pentagonal  prism,  whose  length  is  14 
feet,  and  each  side  of  whose  base  is  33  inches.       Ans.  218.52  sq.  ft. 

3.  Required  the  surface  of  a  cylinder  13  feet  long,  the  circumfer- 
ence of  whose  base  is  57  inches.  Ans.  65.34  square  feet. 

4.  How  often  must  a  cylinder,  5  feet  3  inches  long,  whose  diameter 
is  21  inches,  revolve,  to  roll  an  acre  ?  Ans.  1509.18  times. 

5.  Required  the  wall-surface  of  a  square  room,  whose  sides  are  each 
16  feet  long  and  10  feet  high.  Ans.  71^-  square  yards. 

651  •     To  find  the  contents  or  volume  of  a  prism  or  cylinder. 

Multiply  the  area  of  the  base  of  the  given  prism  or  cylinder  by  the 
height. 

Ex.  1.  What  are  the  contents  of  a  triangular  prism,  whose  length  is 
12  feet,  and  each  side  of  whose  base  is  2^  feet  ?      Ans.  32.47-j-  cu.  ft. 

2.  Required  the  volume  of  a  triangular  prism,  whose  length  is  10 
feet,  and  the  three  sides  of  whose  triangular  end  or  base  are  5,  4, 
and  3  feet.  Ans.  60  cu.  ft. 

*3.  How  many  cubic  feet  in  a  block  of  marble,  whose  length  is  3 
feet  2  inches,  breadth  2  feet  8  inches,  and  depth  2  feet  6  inches  ? 

Ans.  21^  cu.  ft. 

4.  What  is  the  volume  of  a  cylinder,  whose  length  is  9  feet,  and  the 
circumference  of  whose  base  is  6  feet  ?  Ans.  25.78-(-  cu.  ft. 

Pyramids  and  Cones. 

652 •  A  'pyramid  is  a  solid  having  for  its  base  some 
rectilineal  figure,  and  for  its  sides  triangles  meeting  in  a 
common  point  called  the  vertex ;  as  the  figure  A  B. 

The  slant  height  of  a  pyramid  is  a  line  drawn  from  the 
vertex  to  the  middle  of  one  of  the  sides  of  the  base. 

653  •  A  cone  is  a  solid  having  a  circle  for  its  base,  and  tapering 
uniformly  to  a  point  called  the  vertex. 

The  slant  height  of  a  cone  is  a  line  drawn  from  the  vertex  to  the 
circumference  of  the  base. 

654.  The  altitude  or  height  of  a  pyramid  or  of  a  cone  is  a  line 
drawn  from  the  vertex  perpendicular  to  the  plane  of  the  base. 

D 

655.  The  frustum  of  a  solid  is  the  part  that  remains 
after  cutting  off  the  top  by  a  plane  parallel  to  the  base ;  as 
the  frustum  of  a  cone  C  D. 

656 •     To  find  the  surface  of  a  pyramid  or  of  a  cone. 

Multiply  the  perimeter  or  the  circumference  of  the  base  by  half  of  the 
slant  height,  and  to  the  product  add  the  area  of  the  base. 


MENSURATION.  435 

Ex.  1.  Required  the  area  of  the  surface  of  a  square  pyramid,  whose 
base  is  2  feet  8  inches  square,  and  whose  slant  height  is  3  feet  9  inches. 

2.  What  is  the  convex  surface  of  a  cone,  whose  slant  height  is  20 
feet,  and  the  circumference  of  whose  base  is  9  feet  ?      Ans.  90  feet. 

65  7  •     To  find  the  volume  of  a  pyramid,  or  of  a  cone. 
Multiply  the  area  of  the  base  by  one  third  of  the  altitude. 

Ex.  1.  What  is  the  solidity  of  a  cone,  whose  height  is  12-|  feet, 
and  the  diameter  of  whose  base  is  21  feet  ?  Ans.  20.45-f-  feet. 

2.  What  are  the  contents  of  a  triangular  pyramid,  whose  height  is 
14  feet  6  inches,  and  the  sides  of  whose  base  are  5,  6,  and  7  feet  ? 

Ans.  71.035-f  feet 

65 8  •     To  find  the  surface  of  a  frustum  of  a  pyramid,  or  of  a  cone. 

Multiply  the  sum  of  the  perimeters  or  of  the  circumferences  of  the  two 
ends  by  half  of  the  slant  height;  and  to  the  product  add  the  areas  of  the 
two  ends. 

Ex.  1.  Required  the  surface  of  a  frustum  of  a  pentagonal  pyramid, 
whose  slant  height  is  10  inches,  and  the  sides  of  whose  base  are  3 
and  5  inches. 

2.  What  is  the  surface  of  the  frustum  of  a  cone,  the  diameters  of  the 
bases  being  43  and  23  inches,  and  the  slant  height  9  feet  ? 

Ans.  90.72-f-  sq.  ft. 

659.     To  find  the  volume  of  a  frustum  of  a  pyramid,  or  of  a  cone. 

Multiply  the  areas  of  the  two  ends  together,  and  extract  the  square 
root  of  the  product.  To  this  root  add  the  two  areas,  and  multiply  their 
sum  by  one  third  of  the  altitude. 

Ex.  1.  If  the  length  of  a  frustum  of  a  square  pyramid  be  18  feet 
8  inches,  the  side  of  its  greater  base  27  inches,  and  that  of  its  less  16 
inches,  what  is  the  volume  ?  Ans.  61.228-f-  cu.  ft. 

2.  What  are  the  contents  of  a  stick  of  timber,  whose  length  is  40 
feet,  the  diameter  of  the  larger  end  being  24  inches,  and  of  the  smaller 
end  12  inches  ?  Ans.  73  J  cu.  ft.,  nearly. 

Spheres,  Spheroids,  &c. 

660  •  A  sphere  is  a  solid,  bounded  by  a  curved  surface,  every  part 
of  which  is  equally  distant  from  a  point  within,  called  the  centre. 

A 

The  axis  or  diameter  of  a  sphere  is  a  line  passing 
through  the  centre,  and  terminated  by  the  surface ; 
as  the  line  A  B. 


The  radius  of  a  sphere  is  a  line  drawn  from  the  centre  to  any  part 
of  the  surface. 


436 


MENSURATION. 


661  •  A  segment  of  a  sphere  is  a  part  of  it 
cut  off  by  any  plane ;  as  the  figure  ACBD. 
The  plane  is  the  base  of  the  segment ;  the  per- 
pendicular distance  from  the  centre  of  the  base 
to  the  convex  surface  is  the  height  of  the  seg- 
ment ;  as  CD. 

662.  A  spherical  zone  is  a  part  of  the  surface  of  a  sphere  included 
between  two  parallel  planes,  which  form  its  bases ;  and  the  height  of  a 
spherical  zone  is  the  perpendicular  distance  between  the  planes  form- 
ing its  bases. 


663.  A  cylindrical  ring  is  a  figure  formed 
by  bending  a  cylinder  uniformly  till  the  two 
ends  meet ;  as  A  C  D  B. 


664.  A  spheroid  is  a  figure  resembling 
a  sphere,  and  which  may  be  formed  by  the 
revolution  of  an  ellipse  about  one  of  its  axes ; 
as  AECBDP  A. 

If  the  ellipse  revolves  about  its  longer  or 
transverse  axis  or  diameter,  the  spheroid  is 
prolate,  or  oblong ;  if  about  its  shorter  or  con- 
jugate diameter,  the  spheroid  is  oblate,  or  flat- 
tened. 

665.  A  segment  of  a  spheroid  is  a  part  cut  off  by  any  plane,  as 
F  AEF. 

666.  To  find  the  surface  of  a  sphere. 
Multiply  the  diameter  by  the  circumference. 

Ex.  1.  Required  the  convex  surface  of  a  globe,  whose  diameter  is 
24  inches.  Ans.  1809.55-f-  inches. 

2.  Required  the  surface  of  the  earth,  its  diameter  being  7957J  miles, 
and  its  circumference  25,000  miles.     Ans.  198943750  square  miles. 

667.  To  find  the  solidity  of  a  sphere. 
Multiply  the  cube  of  the  diameter  by  .523598. 

Ex.  1.  What  is  the  solidity  of  a  sphere,  whose  diameter  is  12 
inches  ?  Ans.  904.78+  inches. 

2.  Required  the  solidity  of  the  earth,  supposing  its  circumference  to 
be  25,000  miles.  Ans.  263858149120.06886875  miles. 

668.  To  find  the  convex  surface  of  a  segment  or  of  a  zone  of  a 
sphere. 

Multiply  the  height  of  the  segment  or  zone  by  the  circumference  of  the 
sphere  of  which  it  is  a  part 


MENSURATION.  437 

Ex.  1.     If  the  diameter  of  a  sphere  is  12 \  feet,  what  is  the  convex 
surface  of  a  segment  cut  off  from  it,  whose  height  is  2  feet  ? 

Ans.  78.54  sq.  ft. 
2.  If  the  diameter  of  the  earth,  considered  as  a  perfect  sphere,  is 
7970   miles,  and  the   height  of  each  temperate   zone   is  taken   at 
2143.623553  miles,  what  is  the  surface  of  each  temperate  zone? 

Ans.  53673229.81-f-  sq.  m. 

669  •     To  find  the  solidity  of  a  segment  of  a  sphere. 

Multiply  the  square  of  the  height  plus  three  times  the  square  of  the 
radius  of  the  base,  by  the  height,  and  this  product  by  .5236. 

Ex.  1.     Required  the  solidity  of  a  spherical  segment,  whose  height 
is  3  feet,  and  the  radius  of  whose  base  is  4-|  feet.   Ans.  109.56  cu.  ft. 
.     2.  Required  the  solidity  of  the  segment  of  a  sphere,  whose  height  is 
&  feet,  and  the  diameter  of  whose  base  is  20  feet. 

Ans.  1795.42  cu.  ft. 

670*     To  find  the  surface  of  a  cylindrical  ring. 

Multiply  the  sum  of  the  thickness  and  the  inner  diameter  by  the  thick- 
ness, and  that  product  by  9.8696. 

Ex.  1.    Required  the  surface  of  a  cylindrical  ring,  whose  inner 
diameter  is  21  inches,  and  whose  thickness  is  4  inches. 

Ans.  986.96  sq.  in. 

671  •     To  find  the  solidity  of  a  cylindrical  ring. 

Multiply  the  sum  of  the  thickness  and  the  inner  diameter  by  the  square 
of  the  thickness,  and  that  product  by  2.4674. 

Ex.  1.    Required  the  solidity  of  a  cylindrical  ring,  whose  inner 
diameter  is  25  inches,  and  whose  thickness  is  5  inches. 

Ans.  1850.55  cu.  in. 

672  •     To  find  the  solidity  of  a  spheroid. 

Multiply  the  square  of  the  revolving  axis  by  the  fixed  axis,  and  that 
product  by  .523598. 

Ex.  1.     If  the  fixed  axis  of  a  spheroid  is  32  inches,  and  the  revolv- 
ing axis  20  inches,  what  is  the  solidity  ?  Ans.  6702.08  cu.  in. 

2.  Required  the  contents  of  a  balloon  in  the  form  of  a  prolate 
spheroid,  having  its  longer  diameter  48  feet,  and  its  shorter  38  feet. 

Ans.  36291.76  cu.  ft. 


MENSURATION  OF  LUMBER. 

673*     Boards  are  usually  measured  by  the  square  foot. 

Planks,  joists,  beams,  &c.  are  usually  measured  by  board  measure, 
the  board  being  considered  to  be  1  inch  in  thickness. 

Round  timber  is  sometimes  measured  by  the  ton,  and  sometimes  by 
board  measure. 

37* 


438  MENSURATION. 

674.  To  find  the  number  of  square  feet  in  a  board. 

Multiply  tlie  length  of  the  board,  taken  in  feet,  by  its  width,  taken  in 
inches ,  and  the  product  divided  by  12  will  give  the  contents  in  square 
feet.     Or, 

Take  both  the  length  and  width  in  feet,  and  their  product  will  be  the 
contents  in  feet 

Note.  —  If  the  board  is  tapering,  take  half  the  sum  of  the  width  of  its  ends 
for  the  width. 

Ex.  1.  What  are  the  contents  of  a  board  2-4  feet  long,  and  8 
inches  wide  ?  Ans.  10  feet 

2.  What  are  the  contents  of  a  board  30  feet  long,  and  16  inches 
wide  ?  Ans.  40  feet. 

3.  What  are  the  contents  of  a  tapering  board,  30  feet  long,  whose 
ends  are,  the  one  26  inches,  and  the  other  14  inches  wide  ? 

675.  To  find  the  number  of  feet,  board  measure,  in  a  plank,  joist, 
beam,  &c. 

Multiply  the  ividth  taken  in  inches  by  the  thickness  in  inches,  and  this 
product  by  the  length,  in  feet ;  and  the  last  product  divided  by  12  icill 
give  the  contents  in  feet,  board  measure. 

Note.  — If  the  plank,  joist,  &c.  is  tapering  in  width,  take  half  the  sum  of 
the  width  of  the  ends  for  the  width ;  and  if  the  taper  be  both  of  the  width  and 
the  thickness,  the  common  rule  of  obtaining  the  contents  in  cubic  feet  is,  to 
multiply  half  the  sum  of  the  areas  of  the  two  ends  by  the  length,  and  divide  the 
product  by  144. 

Ex.  1.  How  many  feet  are  there  in  3  joists,  which  are  15  feet  long, 
5  inches  wide,  and  3  inches  thick  ?  Ans.  56 \  feet 

2.  How  many  feet  in  20  joists,  10  feet  long,  6  inches  wide,  and  2 
inches  thick  ?  Ans.  200  feet. 

3.  How  many  feet  in  a  beam  20  feet  long,  10  inches  thick,  whose 
width  tapers  from  18  to  16  inches  ?  Ans.  283  J  feet. 

676 •     To  find  the  contents  of  round  timber. 

Multiply  the  length,  taken  in  feet,  by  the  square  of  one  fourth  of  the 
mean  girth,  taken  in  inches ,  and  this  product  divided  by  144  will  give 
the  contents  in  cubic  feet. 

Note.  —  The  girth  of  tapering  timber  is  usually  taken  about  £  the  distance 
from  the  larger  to  the  smaller  end. 

The  rule  is  that  in  common  use,  though  very  far  from  giving  the  actual 
number  of  cubic  feet  in  round  lumber  measured  by  it.  40  cubic  feet,  as  given 
by  the  rule,  are  in  fact  equal  to  50^  true  cubic  feet.  The  following  rule 
gives  results  more  nearly  accurate,  requiring  to  be  diminished  by  only  one 
foot  in  190,  to  give  exact  contents.  Multiply  the  square  of  one  fifth  of  the 
mean  girth,  taken  in  inches,  by  twice  the  length,  in  feet ;  and  divide  by  144. 

Ex.  1.  How  many  cubic  feet  in  a  stick  of  timber  which  is  30  feet 
long,  and  whose  girth  is  40  inches  ?  Ans.  20f  feet. 

2.  If  a  stick  of  timber  is  50  feet  long,  and  its  girth  is  56  inches, 
what  number  of  cubic  feet  does  it  contain  ?  Ans.  68^-g  feet. 

3.  What  are  the  contents  of  a  log  90  feet  long,  and  whose  circum- 
ference is  120  inches  ?  Ans.  562^  feet. 


MENSURATION.  439 


GAUGING  OF  CASKS. 

677  •  Gauging  is  the  process  of  finding  the  capacities  of  casks  or 
other  vessels. 

Casks  are  generally  considered  to  be  of  four  varieties  :  1.  Having 
the  staves  nearly  straight ;  2 ,  Having  the  staves  very  little  curved  ; 
3,  Having  the  staves  of  a  medium  curve  ;  4.  Having  the  staves  con- 
siderably curved. 

Note.  —  Casks  of  the  first  variety  approach  very  nearly  the  form  of  a 
cylinder ;  those  of  the  third  variety  are  of  the  shape  of  a  molasses  hogshead ; 
those  of  the  second  variety  have  a  curvature  of  stave  between  that  of  the  first 
and  third ;  and  the  fourth  have  a  greater  curvature  than  that  of  the  third. 

678.  In  gauging  casks,  it  is  necessary  first  to  find  the  mean  di- 
ameter. This  is  found  by  taking  the  end  and  middle  diameters,  and 
the  length  in  inches ;  and  then  adding  to  the  end  diameter  the  pro- 
duct of  the  difference  between  the  end  and  middle  diameters  by  .55, 
.60,  .65,  or  .70,  as  the  cask  may  be  of  the  first,  second,  third,  or  fourth, 
variety. 

679*     To  find  the  capacity  of  a  cask  in  gallons, 

Multiply  the  square  of  the  mean  diameter,  in  inches,  by  the  length,  in 
inches  ;  and  the  product  multiplied  by  .0034  will  give  the  capacity  in 
liquid  or  wine  gallons. 

Note  1.  — If  the  capacity  is  required  in  ale  or  beer  gallons,  use  for  a  multi- 
plier .0028  instead  of  .0034.  If  imperial  gallons  are  required;  multiply  the 
liquid  or  wine  gallon,  as  found  by  the  rule,  by  .833. 

Note  2.  —  The  contents  of  any  vessel  being  known  in  cubic  inches,  its  capa- 
city in  liquid  gallons  may  be  found  by  dividing  by  231 ;  in  ale  or  beer  gallons, 
by  dividing  by  282 ;  and  in  bushels,  by  dividing  by  2150.42. 

Ex.  1.  Required  the  capacity  in  gallons  of  a  cask  of  the  fourth 
variety,  whose  middle  diameter  is  35  inches,  head  diameter  27  inches, 
and  length  45  inches.  Ans.  162.6. 

2.  What  is  the  capacity  in  gallons  of  a  cask  of  the  third  variety, 
whose  middle  diameter  is  38  inches,  head  diameter  30  inches,  and 
length  42  inches  ? 

3.  What  are  the  contents  in  liquid  measure  of  a  tub  40  inches  in 
diameter  at  the  top,  30  inches  at  the  bottom,  and  whose  height  is  50 
inches?  Ans.  209.66gal. 

4.  How  many  wine  gallons  will  a  cubical  box  contain,  that  is  10 
feet  long,  5  feet  wide,  and  4  feet  high  ?  Ans.  1496T8Tgal. 

5.  How  many  ale  gallons  will  a  trough  contain,  that  is  1 2  feet  long, 
6  feet  wide,  and  2Jeet  high  ?  Ans.  88 2  Jf  gal. 

6.  How  many  bushels  of  grain  will  a  box  contain  that  is  15  feet 
long,  5  feet  wide,  and  7  feet  high  ?  Ans.  421. 8bu. 

TONNAGE  OF  VESSELS. 

680  •  The  tonnage  of  a  ship  is  the  number  of  tons  burden  it  will 
carry,  with  safety,  under  the  ordinary  circumstances  of  navigation. 


440  MENSURATION. 

The  liglit-loaded  water-line  of  a  vessel  is  the  line  made  by  the  water 
upon  the  outside  of  the  hull  as  it  floats  without  load ;  and  the  deep- 
loaded  water-line  is  that  made  in  like  manner  when  it  is  fully  laden. 

The  number  of  cubic  feet  of  the  hull  between  these  two  water- 
lines,  divided  by  35,  the  number  of  cubic  feet  of  sea-water  which 
must  be  taken  to  weigh  a  ton,  represents  the  weight  of  water  dis- 
placed in  sinking  the  vessel  from  the  light  to  the  deep-loaded  water- 
line,  and  therelbre  its  true  tonnage. 

681.  Government  has  by  law  established  a  rule  by  which  the 
custom-house  officers  are  to  be  guided  in  collecting  tonnage  duties. 
But  as  it  does  not  always  give  the  actual  tonnage,  builders,  and  oth- 
ers, usually  make  their  estimates  by  some  other  rule. 

Government  Rule. 

For  Single-decked  Vessels.  Tale  the  length  on  deck  from 
the  forward  side  of  the  main  stem  to  the  after  side  of  the  stern  post,  and 
the  breadth  at  the  broadest  part  above  the  main  wales ;  take  the  depth 
from  the  under  side  of  the  deck  plank  to  the  ceiling  of  the  hold ;  and 
deduct  from  the  length  three  fifths  of  the  breadth  ;  7iiultij)ly  the  re- 
mainder  by  the  breadth,  and  the  product  by  the  depth ;  and  divide  the 
last  product  by  95. 

For  Double-decked  Vessels.  Proceed  as  with  single-decked 
vessels,  except  for  the  depth  take  half  the  breadth. 

Note.  —  The  government  rule  is  differently  construed.  The  length  is 
usually  taken  in  a  line  with  the  deck;  the  depth  at  the  main  hatch.  But  with 
regard  to  the  breadth,  there  is  a  great  want  of  uniformity  among  measurers; 
most  take  the  breadth  about  45  inches  below  the  plank-sheer  at  the  broadest 
part;  some  consider  the  upper  wale,  and  others  the  lower,  as  the  main  wale, 
thus  making  a  considerable  difference  in  their  results. 

The  government  rule  for  single-decked  vessels  operates  very  well,  but  the 
rule  for  double-decked  vessels,  which  is  also  intended  to  include  all  vessels  of 
more  than  one  deck,  often  fails  to  give  the  true  tonnage.  A  more  accurate 
method  would,  for  double-decked  vessels,  take  the  breadth  5  feet  below  the 
upper  deck,  at  the  broadest  part,  and  for  three-decked  vessels  7  feet  below 
the  upper  deck;  and  in  each  case  for  depth  of  hold  three  ffths  of  the  breadth. 

Ex.  1.  A.  &  G.  T  Sampson,  of  East  Boston,  have  contracted  to 
build  a  clipper  ship  191T6^  feet  lon< 


what  is  the  government  tonnage 

2.  What  is  the  government  tonnage  of  the  ship 
length  is  184T<W,  width  3811,  and  depth  28  feet  ? 

Ans.  1284ffff£7  tons. 

3.  The  ship  Mattakeeset  is  195T%  feet  long,  39^  wide,  and  27^ 
deep ;  what  is  the  government  tonnage  of  the  same  V 

Ans.  13970!^  tons.  ^ 

4.  Required  the  tonnage  of  a  single-decked  vessel,  whose  length  is 
78  feet,  width  21  feet,  and  depth  9  feet  Ans.  130A\  tons. 

5.  What  is  the  government  tonnage  of  a  double-decked  vessel, 
whose  length  is  159  feet,  and  width  30lfeet  ?  Ans.  6671^  tons.^ 

6.  What  is  the  government  tonnage  of  Noah's  ark,  admitting  its 
length  to  have  been  479  feet,  its  breadth  80  feet,  and  its  depth  48 
fee??  Ans.  1451 7lJ  tons. 


MISCELLANEOUS   EXAMPLES.  441 


MISCELLANEOUS    EXAMPLES. 

1.  What  number  is  that  to  which,  if  -f  of  -f  be  added,  the  sum  will 
be  1  ?  Ans.  £§. 

2.  A  certain  gentleman,  at  the  time  of  his  marriage,  agreed  to  give 
his  wife  §  of  his  estate,  if,  at  the  time  of  his  death,  he  left  only  a 
daughter,  and  if  he  left  only  a  son,  she  should  have  -J-  of  his  property : 
but  as  it  happened,  he  left  a  son  and  a  daughter,  in  consequence  or 
which  the  widow  received  in  equity  $  2400  less  than  she  would  have 
received  if  there  had  been  only  a  daughter.  What  would  have  been 
his  wife's  dowry  if  he  had  left  only  a  son  ?  Ans.  $  2100. 

3.  A  gentleman  being  asked  what  o'clock  it  was,  said  that  it  was 
between  5  and  6  ;  but,  to  be  more  particular,  he  said  that  the  minute- 
hand  had  passed  as  far  beyond  the  6  as  the  hour-hand  wanted  of 
having  reached  the  6  ;  that  is,  that  the  hour  and  minute-hands  made 
equal  acute  angles  with  a  line  passing  from  the  12  through  the  6. 
Required  the  time  of  day.  Ans.  32m.  18-j^-s.  past  5. 

4.  Divide  97deg.  55m.  7fur.  35rd.  4ft.  6in.  by  6. 

5.  A,  B,  and  C  are  to  share  $  100,000  in  the  proportion  of  -J,  -J-, 
and  1,  respectively ;  but  C's  part  being  lost  by  his  death,  it  is  required 
to  divide  the  whole  sum  properly  between  the  other  two. 

Ans.  As  part  is  $57,142$,  and  B's  $42,857f 

6.  A  father  devised  T7g-  of  his  estate  to  one  of  his  sons,  and  T7g-  of 
the  residue  to  the  other,  and  the  remainder  to  his  wife.  The  differ- 
ence of  his  sons'  legacies  was  found  to  be  25  7£.  3s.  4d.  What  money 
did  he  leave  for  his  widow?  Ans.  635£.  0s.  10-JJd. 

7.  In  the  walls  of  Balbec,  in  Turkey,  the  ancient  Heliopolis,  there 
are  three  stones  laid  end  to  end,  now  in  sight,  that  measure  61  yards 
in  length,  one  of  which  is  63  feet  long,  12  feet  thick,  and  12  feet 
broad ;  what  is  its  weight,  supposing  its  specific  gravity  to  be  3  times 
that  of  water?  Ans.  75 9|  tons. 

8.  A  burden  of  2001b.,  suspended  on  a  pole  4ft.  in  length,  the 
point  of  suspension  being  6in.  from  the  middle,  is  carried  by  two  men, 
the  ends  of  the  pole  resting  on  their  shoulders  ;  how  much  of  this  load 
is  borne  by  each  man  ?  Ans.  1251b.  and  75lb. 

9.  The  court-house  in  Boston  has  eight  pillars  of  granite,  each 
25ft.  4in.  in  length,  4ft.  5in.  in  diameter  at  one  end,  and  3ft.  5in. 
in  diameter  at  the  other  end.  How  many  cubic  feet  do  they  con- 
tain, and  what  is  their  weight,  allowing  a  cubic  foot  to  weigh  3000 
ounces  ?  Ans.  2455.03  cubic  feet,  205.4  tons. 

10.  A  father,  dying,  left  his  son  a  legacy,  \  of  which  he  spent  in  8 
months  ;  -|  of  the  remainder  lasted  him  1 2  months  longer,  after  which 
he  had  only  $410  left.  What  amount  did  his  father  bequeath 
him  ?  Ans.  $  956.66$. 

11.  A  merchant  sold  goods  to  a  certain  amount,  on  a  commission 
of  4  per  cent.,  and,  having  remitted  the  net  proceeds  to  the  owner, 
received  ^  per  cent,  for  prompt  payment,  which  amounted  to  $  15.60. 
What  was  the  amount  of  his  commission  ?  Ans.  $  260. 


442  MISCELLANEOUS   EXAMPLES. 

12.  A,  of  Boston,  remits  to  B,  of  New  York,  a  bill  of  exchange  on 
London,  the  avails  of  which  he  wishes  to  be  invested  in  goods  on  his 
account.  B,  having  disposed  of  the  bill  at  1\  per  cent,  advance,  re- 
ceived $9675  ;  and  having  reserved  for  himself  \  per  cent,  on  the 
sale  of  the  bill,  and  2  per  cent,  for  commission,  he  invests  the  remain- 
der. What  is  the  amount  invested,  and  for  how  much  was  the  bill 
drawn  ?  Ans.  Investment,  $  9461.58^- ;  the  bill  was  £  2025. 

13.  Bunker  Hill  Monument  is  30ft.  square  at  its  base,  and  15ft. 
square  at  its  top  ;  its  height  is  220  feet.  From  the  bottom  to  the  top, 
through  its  centre,  is  a  conical  avenue  15ft.  in  diameter  at  the  bottom, 
and  lift,  at  the  top.  How  many  cubic  feet  are  there  in  the  monu- 
ment ?  Ans.  86,068.518-f-  ft. 

14.  A  hired  a  house  for  one  year  for  $  300  ;  at  the  end  of  4  months 
he  takes  in  M  as  a  partner,  and  at  the  end  of  8  months  he  takes  in  P. 
At  the  end  of  the  year,  what  rent  must  each  pay  ? 

Ans.  A  pays  $  183 J ;  M  pays  $  83J  ;  P  pays  $  33|. 

15.  A  merchant  receives  on  commission  three  kinds  of  flour ;  from 
A  he  receives  20  barrels,  from  B  25  barrels,  and  'from  C  40  barrels. 
He  finds  that  A's  flour  is  10  per  cent  better  than  B's,  and  that  B's  is 
20  per  cent,  better  than  C's.  He  sells  the  whole  at  $  6  per  barrel. 
What  in  justice  should  each  man  receive  ? 

Ans.  A  receives  $  139JJ $ ;  B,  $  158£J?  ;  C,  211JJ f. 

16.  Bought  100  barrels  of  flour,  at  $5  per  barrel,  and  immediately 
sold  it  on  a  credit  of  six  months.  The  note  which  I  received  for  pay 
I  got  discounted  at  the  Suffolk  Bank,  and,  on  examining  my  money, 
J  round  that  I  had  gained  20  per  cent,  on  my  purchase.  What  did  I 
receive  per  barrel  fbr  the  flour  ?  Ans.  $  6.18^|£§. 

17.  Required  the  greatest  possible  number  of  hills  of  corn  that  can  be 
planted  on  a  square  acre,  the  hills  to  occupy  only  a  mathematical  point, 
and  no  two  hills  to  be  nearer  than  three  and  a  half  feet.     Ans.  4165. 

18.  Lent  a  friend  $  700,  which  he  kept  20  months.  Some  years 
after  I  borrowed  of  him  $  300  ;  how  long  should  I  keep  it  to  balance 
the  favor  ?  Ans.  46§  months. 

19.  John  Lee  gave  \  of  his  estate  to  his  wife,  \  of  the  remainder  to 
his  oldest  son,  and  \  of  the  residue  to  his  oldest  daughter,  and  \  of 
what  then  remained,  which  was  $  1500,  was  to  be  equally  distributed 
among  his  other  children,  who  received  $150  each;  required  the 
number  of  his  children,  and  the  value  of  his  estate. 

20.  A  and  B  set  out  to  travel  round  a  certain  island,  which  is  80 
miles  in  circumference.  A  travels  5  miles  a  day,  and  B  7  miles  a 
day.     How  far  must  B  travel  to  overtake  A?  Ans.  280  miles. 

21.  If  24.4  cubic  inches  of  lead  weigh  16  pounds,  required  the 
number  of  feet  of  lead  pipe  that  can  be  made  from  80  pounds  of  lead, 
the  diameter  of  the  pipe  to  be  1  inch,  and  the  thickness  of  it  \  of  an 
inch.  Ans.  124.26-f-  in. 

22.  How  long  a  tube  can  be  made  from  a  cylinder  of  lead  8  inches 
lon<*  and  2  inches  in  diameter,  and  through  the  centre  of  which  is  a 
hole  4  of  an  inch  in  diameter ;  the  tube  or  pipe  to  be  \  of  an  inch  in 
diameter,  and  \  of  an  inch  in  thickness  ?  Ans.  16.29-[-  in. 

23.  Four  men,  A,  B,  C,  and  D,  bought  a  stack  of  hay,  containing 


MISCELLANEOUS   EXAMPLES.  443 

8  tons,  for  $  100.  A  is  to  have  12  per  cent,  more  of  the  hay  than  B, 
B  is  to  have  10  per  cent,  more  than  C,  and  C  is  to  have  8  per 
cent,  more  than  D.  Each  man  is  to  pay  in  proportion  to  the  quan- 
tity he  receives.  The  stack  is  20  feet  high,  and  12  feet  square  at  its 
base,  it  being  an  exact  pyramid ;  and  it  is  agreed  that  A  shall  take 
his  share  first  from  the  top  of  the  stack,  B  is  to  take  his  share  the 
next,  and  then  C  and  D.  How  many  feet  of  the  perpendicular 
height  of  the  stack  shall  each  take,  and  what  sum  shall  each  pay  ? 
Ans.  A  takes  13.22+ft,  and  pays  $  28.93-|f ff*  ;  B  takes  3.14-fft, 

and  pays  $  25.83flff  *  '■>  C  takes  2.06+ft.,  and  pays  $  23.48^f  if  f ; 

D  takes  1.58-fft,  and  pays  $  21.74i|-0ff .  ^ 

24.  A,  B,  and  C  bought  a  grindstone"  for  which  they  paid  $  10.60. 
B  paid  20  per  cent,  more  than  A,  and  10  per  cent,  less  than  C.  The 
diameter  of  the  stone  was  65  inches,  and  the  diameter  of  the  place  for 
the  shaft  3  inches.  What  sum  did  each  pay,  and  how  much  must 
each  grind  off  from  the  semidiameter  to  obtain  his  proper  share  of 
the  stone  ? 

Ans.  A  paid  $  3,  B  $  3.60,  and  C$4.     A  grinds  off  5  inches  ;  B 
1\  inches,  and  C  18-1-  inches. 

25.  A  servant  draws  off  a  gallon  on  each  day,  for  20  days,  from  a 
cask  containing  10  gallons  of  wine,  each  time  supplying  the  deficiency 
by  the  addition  of  a  gallon  of  water ;  and  then,  to  escape  detection, 
he  again  draws  off  20  gallons,  supplying  the  deficiency  each  time  by  a 
gallon  of  wine.     How  much  water  still  remains  in  the  cask  ? 

Ans.  1.0679577  gallons,  or  more  than  a  gallon  and  half  a  pint. 

26.  The  dimensions  of  a  bushel  measure  are  18-1-  inches  wide,  and  8 
inches  deep ;  what  should  be  the  dimensions  of  a  similar  measure  that 
would  contain  8  bushels?  Ans.  37in.  wide,  16in.  deep. 

27.  What  is  the  weight  of  a  hollow  spherical  iron  shell  5  inches  in 
diameter,  the  thickness  of  the  metal  being  1  inch,  and  a  cubic  inch  of 
iron  weighing  iff  of  a  pound  ?  Ans.  13.2387lb. 

28.  At  a  certain  time  between  2  and  3  o'clock,  the  minute-hand 
was  between  3  and  4.  Within  an  hour  after,  the  hour-hand  and 
minute-hand  had  exactly  changed  places  with  each  other.  What 
was  the  precise  time  when  the  hands  were  in  the  first  position? 

Ans.  2h.  15m.  56T9T\s. 

29.  Required  the  contents  of  the  largest  cube  that  can  be  inscribed 
in  a  sphere  20  inches  in  diameter.  Ans.  1539.58-]-  cu.  in. 

30.  If  in  a  pair  of  scales  a  body  weigh  90  pounds  in  one  scale,  and 
only  40  pounds  in  the  other,  required  the  true  weight,  and  the  propor- 
tion of  the  lengths  of  the  two  arms  of  the  balance-beam  on  each  side 
of  the  point  of  suspension. 

Ans.  Weight  60lb.,  and  the  proportions  3  to  2. 

31.  In  turning  a  one-horse  chaise  within  a  ring  of  a  certain  diame- 
ter, it  was  observed  that  the  outer  wheel  made  two  turns,  while  the 
inner  wheel  made  but  one ;  the  wheels  were  each  4  feet  high ;  and 
supposing  them  fixed  at  the  distance  of  5  feet  asunder  on  the  axletree, 
what  was  the  circumference  of  the  track  described  by  the  outer 
wheel  ?  Ans.  62.83-)-  feet. 

32.  The  ball  on  the  top  of  St.  Paul's  Church  is  6  feet  in  diameter. 
What  did  the  gilding  of  it  cost,  at  8-Jd.  per  square  inch  V 


444  MISCELLANEOUS   EXAMPLES. 

33.  There  is  a  conical  glass,  6  inches  high,  5  inches  wide  at  the  top, 
and  which  is  ^  part  filled  with  water.  What  must  be  the  diameter  of 
a  ball,  let  fall  into  the  water,  that  shall  be  immersed  by  it  ? 

Ans.  2.44  5-|-  inches. 

34.  A  certain  lady,  the  mother  of  three  daughters,  had  a  farm  of 
500  acres,  in  a  circular  form,  with  her  dwelling-house  in  the  centre. 
Being  desirous  of  having  her  daughters  settled  near  her,  she  gave  to 
them  three  equal  parcels,  as  large  as  could  be  made  in  three  equal 
circles  included  within  the  periphery  of  her  farm,  one  to  each,  with  a 
dwelling-house  in  the  centre  of  each;  that  is,  there  were  to  be  three 
equal  circles,  as  large  as  could  be  made  within  a  circle  that  contained 
500  acres.  How  many  acres  did  the  fann  of  each  daughter  contain, 
how  many  acres  did  the  mother  retain,  how  far  apart  were  the  dwell- 
ing-houses of  the  daughters,  and  how  far  was  the  dwelling-house  of 
each  daughter  from  that  of  the  mother? 

Ans.  Each  daughter's  fann  contained  107  acres  2  roods  31.22-j— 
rods.  The  mother  retained  176  acres  3  roods  26.34-}- rods.  The 
distance  from  one  daughter's  house  to  the  other  was  148.1 1981 7-{- 
rods.  The  mother's  dwelling-house  was  distant  from  her  daugh- 
ters' 85.5 1-\-  rods. 

35.  James  Page  has  a  circular  garden,  10  rods  in  diameter ;  how 
many  frees  can  be  set  in  it,  so  that  no  two  shall  be  within  ten  feet 
of  each  other,  and  no  tree  within  two  and  a  halt'  feet  of  the  fence  en- 
closing the  garden  ?  Ans.  241. 

36.  A  and  B  engaged  to  reap  a  field  for  90  shillings ;  and  as  A 
could  reap  it  in  9  days,  they  promised  to  complete  it  in  5  days.  They 
found,  however,  that  they  were  obliged  to  call  in  C,  an  inferior  work- 
man, to  assist  them  for  the  last  two  days,  in  consequence  of  which  B 
received  3s.  9d.  less  than  lie  otherwise  would  have  done.  In  what 
time  could  B  and  G  each  reap  the  iield  ? 

Ans.  B  could  reap  it  in  15  days,  and  C  in  18  days. 

37.  A  merchant  tailor  bought  40  yards  of  broadcloth,  2-J-  yards 
wide  ;  but  on  sponging  it,  it  shrunk  in  length  upon  every  4  yards 
half  a  quarter,  and  in  width,  one  nail  and  a  half  upon  every  1^  yards. 
To  line  this  cloth,  he  bought  flannel  5  quarters  wide,  which,  being  wet, 
shrunk  the  whole  width  on  every  20  yards  in  length,  and  in  width  it 
shrunk  half  a  nail.  Required  the  number  of  yards  of  flannel  used  in 
lining  the  cloth.  Ans.  71T7T  yards. 

38.  1  have  a  garden  in  the  form  of  an  equilateral  triangle,  whose 
sides  are  200  feet.  At  each  corner  stands  a  tower  ;  the  height  of  the 
fust  is  30  feet,  the  second  40,  and  the  third  50.  At  what  distance 
from  the  base  of  each  tower  must  a  ladder  be  placed,  that  it  may  just 
reach  the  top  of  each  ?  And  what  is  the  length  of  the  ladder,  the 
garden  being  a  horizontal  plane  ? 

'  Ans.  The  "foot  of  the  ladder  from  the  base  of  the  first  tower  118.811-}- 
feet;  second  tower,  1 1 5.82 7-f-  feet ;  third  tower,  111.875-)- feet. 
Length  of  the  ladder,  1 22.535-f-  feet. 


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